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The Quadratic Formula can be daunting but this link from Khan Academy breaks it down a bit more if you need extra information.  It also includes an interesting video about proving the Quadratic Formula.
https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-using-the-quadratic-formula/a/quadratic-formula-explained-article
Source: Khan, S. (2010). Understanding the quadratic formula. Retrieved May 08, 2017, from https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-using-the-quadratic-formula/a/quadratic-formula-explained-article

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In this video, you will see a few examples of quadratic equations being solved using the quadratic formula.
Source: Sal Khan "The Quadratic Formula."  https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-using-the-quadratic-formula/v/using-the-quadratic-formula  Khan Academy, 12 April, 2010.  Retrieved 8 May, 2017.
CC-BY-NC-SA 3.0

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In this video, I'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics. And if you've seen many of my videos, you know that I'm not a big fan of memorizing things. But I will recommend you memorize it with the caveat that you also remember how to prove it, because I don't want you to just remember things and not know where they came from. But with that said, let me show you what I'm talking about: it's the quadratic formula. And as you might guess, it is to solve for the roots, or the zeroes of quadratic equations. So let's speak in very general terms and I'll show you some examples. So let's say I have an equation of the form ax squared plus bx plus c is equal to 0. You should recognize this. This is a quadratic equation where a, b and c are-- Well, a is the coefficient on the x squared term or the second degree term, b is the coefficient on the x term and then c, is, you could imagine, the coefficient on the x to the zero term, or it's the constant term. Now, given that you have a general quadratic equation like this, the quadratic formula tells us that the solutions to this equation are x is equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. And I know it seems crazy and convoluted and hard for you to memorize right now, but as you get a lot more practice you'll see that it actually is a pretty reasonable formula to stick in your brain someplace. And you might say, gee, this is a wacky formula, where did it come from? And in the next video I'm going to show you where it came from. But I want you to get used to using it first. But it really just came from completing the square on this equation right there. If you complete the square here, you're actually going to get this solution and that is the quadratic formula, right there. So let's apply it to some problems. Let's start off with something that we could have factored just to verify that it's giving us the same answer. So let's say we have x squared plus 4x minus 21 is equal to 0. So in this situation-- let me do that in a different color --a is equal to 1, right? The coefficient on the x squared term is 1. b is equal to 4, the coefficient on the x-term. And then c is equal to negative 21, the constant term. And let's just plug it in the formula, so what do we get? We get x, this tells us that x is going to be equal to negative b. Negative b is negative 4-- I put the negative sign in front of that --negative b plus or minus the square root of b squared. b squared is 16, right? 4 squared is 16, minus 4 times a, which is 1, times c, which is negative 21. So we can put a 21 out there and that negative sign will cancel out just like that with that-- Since this is the first time we're doing it, let me not skip too many steps. So negative 21, just so you can see how it fit in, and then all of that over 2a. a is 1, so all of that over 2. So what does this simplify, or hopefully it simplifies? So we get x is equal to negative 4 plus or minus the square root of-- Let's see we have a negative times a negative, that's going to give us a positive. And we had 16 plus, let's see this is 6, 4 times 1 is 4 times 21 is 84. 16 plus 84 is 100. That's nice. That's a nice perfect square. All of that over 2, and so this is going to be equal to negative 4 plus or minus 10 over 2. We could just divide both of these terms by 2 right now. So this is equal to negative 4 divided by 2 is negative 2 plus or minus 10 divided by 2 is 5. So that tells us that x could be equal to negative 2 plus 5, which is 3, or x could be equal to negative 2 minus 5, which is negative 7. So the quadratic formula seems to have given us an answer for this. You can verify just by substituting back in that these do work, or you could even just try to factor this right here. You say what two numbers when you take their product, you get negative 21 and when you take their sum you get positive 4? So you'd get x plus 7 times x minus 3 is equal to negative 21. Notice 7 times negative 3 is negative 21, 7 minus 3 is positive 4. You would get x plus-- sorry it's not negative --21 is equal to 0. There should be a 0 there. So you get x plus 7 is equal to 0, or x minus 3 is equal to 0. X could be equal to negative 7 or x could be equal to 3. So it definitely gives us the same answer as factoring, so you might say, hey why bother with this crazy mess? And the reason we want to bother with this crazy mess is it'll also work for problems that are hard to factor. And let's do a couple of those, let's do some hard-to-factor problems right now. So let's scroll down to get some fresh real estate. Let's rewrite the formula again, just in case we haven't had it memorized yet. x is going to be equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. I'll supply this to another problem. Let's say we have the equation 3x squared plus 6x is equal to negative 10. Well, the first thing we want to do is get it in the form where all of our terms or on the left-hand side, so let's add 10 to both sides of this equation. We get 3x squared plus the 6x plus 10 is equal to 0. And now we can use a quadratic formula. So let's apply it here. So a is equal to 3. That is a, this is b and this right here is c. So the quadratic formula tells us the solutions to this equation. The roots of this quadratic function, I guess we could call it. x is going to be equal to negative b. b is 6, so negative 6 plus or minus the square root of b squared. b is 6, so we get 6 squared minus 4 times a, which is 3 times c, which is 10. Let's stretch out the radical little bit, all of that over 2 times a, 2 times 3. So we get x is equal to negative 6 plus or minus the square root of 36 minus-- this is interesting --minus 4 times 3 times 10. So this is minus-- 4 times 3 times 10. So this is minus 120. All of that over 6. So this is interesting, you might already realize why it's interesting. What is this going to simplify to? 36 minus 120 is what? That's 84. We make this into a 10, this will become an 11, this is a 4. It is 84, so this is going to be equal to negative 6 plus or minus the square root of-- But not positive 84, that's if it's 120 minus 36. We have 36 minus 120. It's going to be negative 84 all of that 6. So you might say, gee, this is crazy. What a this silly quadratic formula you're introducing me to, Sal? It's worthless. It just gives me a square root of a negative number. It's not giving me an answer. And the reason why it's not giving you an answer, at least an answer that you might want, is because this will have no real solutions. In the future, we're going to introduce something called an imaginary number, which is a square root of a negative number, and then we can actually express this in terms of those numbers. So this actually does have solutions, but they involve imaginary numbers. So this actually has no real solutions, we're taking the square root of a negative number. So the b squared with the b squared minus 4ac, if this term right here is negative, then you're not going to have any real solutions. And let's verify that for ourselves. Let's get our graphic calculator out and let's graph this equation right here. So, let's get the graphs that y is equal to-- that's what I had there before --3x squared plus 6x plus 10. So that's the equation and we're going to see where it intersects the x-axis. Where does it equal 0? So let me graph it. Notice, this thing just comes down and then goes back up. Its vertex is sitting here above the x-axis and it's upward-opening. It never intersects the x-axis. So at no point will this expression, will this function, equal 0. At no point will y equal 0 on this graph. So once again, the quadratic formula seems to be working. Let's do one more example, you can never see enough examples here. And I want to do ones that are, you know, maybe not so obvious to factor. So let's say we get negative 3x squared plus 12x plus 1 is equal to 0. Now let's try to do it just having the quadratic formula in our brain. So the x's that satisfy this equation are going to be negative b. This is b So negative b is negative 12 plus or minus the square root of b squared, of 144, that's b squared minus 4 times a, which is negative 3 times c, which is 1, all of that over 2 times a, over 2 times negative 3. So all of that over negative 6, this is going to be equal to negative 12 plus or minus the square root of-- What is this? It's a negative times a negative so they cancel out. So I have 144 plus 12, so that is 156, right? 144 plus 12, all of that over negative 6. Now, I suspect we can simplify this 156. We could maybe bring some things out of the radical sign. So let's attempt to do that. So let's do a prime factorization of 156. Sometimes, this is the hardest part, simplifying the radical. So 156 is the same thing as 2 times 78. 78 is the same thing as 2 times what? That's 2 times 39. So the square root of 156 is equal to the square root of 2 times 2 times 39 or we could say that's the square root of 2 times 2 times the square root of 39. And this, obviously, is just going to be the square root of 4 or this is the square root of 2 times 2 is just 2. 2 square roots of 39, if I did that properly, let's see, 4 times 39. Yeah, it looks like it's right. So this up here will simplify to negative 12 plus or minus 2 times the square root of 39, all of that over negative 6. Now we can divide the numerator and the denominator maybe by 2. So this will be equal to negative 6 plus or minus the square root of 39 over negative 3. Or we could separate these two terms out. We could say this is equal to negative 6 over negative 3 plus or minus the square root of 39 over negative 3. Now, this is just a 2 right here, right? These cancel out, 6 divided by 3 is 2, so we get 2. And now notice, if this is plus and we use this minus sign, the plus will become negative and the negative will become positive. But it still doesn't matter, right? We could say minus or plus, that's the same thing as plus or minus the square root of 39 nine over 3. I think that's about as simple as we can get this answered. I want to make a very clear point of what I did that last step. I did not forget about this negative sign. I just said it doesn't matter. It's going to turn the positive into the negative; it's going to turn the negative into the positive. Let me rewrite this. So this right here can be rewritten as 2 plus the square root of 39 over negative 3 or 2 minus the square root of 39 over negative 3, right? That's what the plus or minus means, it could be this or that or both of them, really. Now in this situation, this negative 3 will turn into 2 minus the square root of 39 over 3, right? I'm just taking this negative out. Here the negative and the negative will become a positive, and you get 2 plus the square root of 39 over 3, right? A negative times a negative is a positive. So once again, you have 2 plus or minus the square of 39 over 3. 2 plus or minus the square root of 39 over 3 are solutions to this equation right there. Let verify. I'm just curious what the graph looks like. So let's just look at it. Let me clear this. Where is the clear button? So we have negative 3 three squared plus 12x plus 1 and let's graph it. Let's see where it intersects the x-axis. It goes up there and then back down again. So 2 plus or minus the square, you see-- The square root of 39 is going to be a little bit more than 6, right? Because 36 is 6 squared. So it's going be a little bit more than 6, so this is going to be a little bit more than 2. A little bit more than 6 divided by 2 is a little bit more than 2. So you're going to get one value that's a little bit more than 4 and then another value that should be a little bit less than 1. And that looks like the case, you have 1, 2, 3, 4. You have a value that's pretty close to 4, and then you have another value that is a little bit-- It looks close to 0 but maybe a little bit less than that. So anyway, hopefully you found this application of the quadratic formula helpful.

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The quadratic formula | Algebra (video) | Khan Academy

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The quadratic formula helps us solve any quadratic equation. First, we bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. Then, we plug these coefficients in the formula: (-b±√(b²-4ac))/(2a) . See examples of using the formula to solve a variety of equations.

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Source: Dane Elhert, "Quadratic Formula."  Open Middle, 2015. http://www.openmiddle.com/wp-content/uploads/2014/11/quadratic_formula.png Retrieved 8 May, 2017.
CC-BY-NC-SA 4.0

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Solving Quadratics by Completing the Square

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Solving Quadratics by Completing the Square

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The following link shows the steps in using the method known as completing the square to solve a quadratic for x.
http://www.openalgebra.com/2012/11/completing-square.html
Source: John Redden "OpenAlgebra.com Completing the Square." Open Algebra, 8 November, 2012.  Retrieved 8 May, 2017.
CC-BY-NC-SA 3.0

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Find the error in the students work when completing the square.
Source: Michael Pershan "Completing the Square."  Math Mistakes, 22 April, 2013.  http://i1.wp.com/mathmistakes.org/wp-content/uploads/2013/04/photo-61.jpg  Retrieved 8 May, 2017
CC-BY 3.0

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In this video, 4 different quadratics are solved by completing the square.
Source: OHSU Teacher "6.5 Solving Quadratic Equations by Completing the Square." Vimeo, 24 June, 2010.  Retrieved 8 May, 2017.
CC-BY 4.0

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6.5 Solving Quadratic Equations by Completing the Square

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This is "6.5 Solving Quadratic Equations by Completing the Square" by OHSU Teacher on Vimeo, the home for high quality videos and the people who love them.

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In the link provided, you will see how we can apply the concept of factoring from our previous unit to solving a quadratic.  Please read over the information and attempt the practice problems provided.
https://courses.lumenlearning.com/collegealgebra1/chapter/solving-quadratic-equations-by-factoring/
Source: OpenStax College Algebra "Solving Quadratic Equations by Factoring." Lumen Learning, 2017. Retrieved 8 May, 2017.
CC-BY 4.0

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Graphing with the Axis of Symmetry and Vertex

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Graphing with the Axis of Symmetry and Vertex

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In this link, you will briefly see how to apply the vertex formula to help graph a parabola.
https://www.mathplanet.com/education/algebra-1/quadratic-equations/the-graph-of-y-ax-2-plus-bx-plus-c
Source: Mathplanet "The graph of y = ax^2 + bx + c."  Mathplanet, 2017.  Retrieved 8 May, 2017.
CC-BY-NC-ND

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In the Khan Academy video, you will learn an alternate method on how to find the Axis of Symmetry and Vertex of a Parabola
Source: Sal Khan "Vertex and Axis of Symmetry of a Parabola."  Khan Academy, 23 June. 2010.  Retrieved 8 May, 2017.
CC-BY-NC-SA 3.0

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We need to find the vertex and the axis of symmetry of this graph. The whole point of doing this problem is so that you understand what the vertex and axis of symmetry is. And just as a bit of a refresher, if a parabola looks like this, the vertex is the lowest point here, so this minimum point here, for an upward opening problem. If the parabola opens downward like this, the vertex is the topmost point right like that. It's the maximum point. And the axis of symmetry is the line that you could reflect the parabola around, and it's symmetric. So that's the axis of symmetry. That is a reflection of the left-hand side along that axis of symmetry. Same thing if it's a downward-opening parabola. And the general way of telling the difference between an upward-opening and a downward-opening parabola is that this will have a positive coefficient on the x squared term, and this will have a negative coefficient. And we'll see that in a little bit more detail. So let's just work on this. Now, in order to figure out the vertex, there's a quick and dirty formula, but I'm not going to do the formula here because the formula really tells you nothing about how you got it. But I'll show you how to apply the formula at the end of this video, if you see this on a math test and just want to do it really quickly. But we're going to do it the slow, intuitive way first. So let's think about how we can find either the maximum or the minimum point of this parabola. So the best way I can think of doing it is to complete the square. And it might seem like a very foreign concept right now, but let's just do it one step at a time. So I can rewrite this as y is equal to-- well, I can factor out a negative 2. It's equal to negative 2 times x squared minus 4x minus 4. And I'm going to put the minus 4 out here. And this is where I'm going to complete the square. Now, what I want to do is express the stuff in the parentheses as a sum of a perfect square and then some number over here. And I have x squared minus 4x. If I wanted this to be a perfect square, it would be a perfect square if I had a positive 4 over here. If I had a positive 4 over there, then this would be a perfect square. It would be x minus 2 squared. And I got the 4, because I said, well, I want whatever half of this number is, so half of negative 4 is negative 2. Let me square it. That'll give me a positive 4 right there. But I can't just add a 4 willy-nilly to one side of an equation. I either have to add it to the other side or I would have to then just subtract it. So here I haven't changed equation. I added 4 and then I subtracted 4. I just added zero to this little expression here, so it didn't change it. But what it does allow me to do is express this part right here as a perfect square. x squared minus 4x plus 4 is x minus 2 squared. It is x minus 2 squared. And then you have this negative 2 out front multiplying everything, and then you have a negative 4 minus negative 4, minus 8, just like that. So you have y is equal to negative 2 times this entire thing, and now we can multiply out the negative 2 again. So we can distribute it. Y is equal to negative 2 times x minus 2 squared. And then negative 2 times negative 8 is plus 16. Now, all I did is algebraically arrange this equation. But what this allows us to do is think about what the maximum or minimum point of this equation is. So let's just explore this a little bit. This quantity right here, x minus 2 squared, if you're squaring anything, this is always going to be a positive quantity. That right there is always positive. But it's being multiplied by a negative number. So if you look at the larger context, if you look at the always positive multiplied by the negative 2, that's going to be always negative. And the more positive that this number becomes when you multiply it by a negative, the more negative this entire expression becomes. So if you think about it, this is going to be a downward-opening parabola. We have a negative coefficient out here. And the maximum point on this downward-opening parabola is when this expression right here is as small as possible. If this gets any larger, it's just multiplied by a negative number, and then you subtract it from 16. So if this expression right here is 0, then we have our maximum y value, which is 16. So how do we get x is equal to 0 here? Well, the way to get x minus 2 equal to 0-- so let's just do it. x minus 2 is equal to 0, so that happens when x is equal to 2. So when x is equal to 2, this expression is 0. 0 times a negative number, it's all 0, and then y is equal to 16. This is our vertex, this is our maximum point. We just reasoned through it, just looking at the algebra, that the highest value this can take on is 16. As x moves away from 2 in the positive or negative direction, this quantity right here, it might be negative or positive, but when you square it, it's going to be positive. And when you multiply it by negative 2, it's going to become negative and it's going to subtract from 16. So our vertex right here is x is equal to 2. Actually, let's say each of these units are 2. So this is 2, 4, 6, 8, 10, 12, 14, 16. So my vertex is here. That is the absolute maximum point for this parabola. And its axis of symmetry is going to be along the line x is equal to 2, along the vertical line x is equal to 2. That is going to be its axis of symmetry. And now if we're just curious for a couple of other points, just because we want to plot this thing, we could say, well, what happens when x is equal to 0? That's an easy one. When x is equal to 0, y is equal to 8. So when x is equal to 0, we have 1, 2, 3, 4-- oh, well, these are 2. 2, 4, 6, 8. It's right there. This is an axis of symmetry. So when x is equal to 3, y is also going to be equal to 8. So this parabola is a really steep and narrow one that looks something like this, where this right here is the maximum point. Now I told you this is the slow and intuitive way to do the problem. If you wanted a quick and dirty way to figure out a vertex, there is a formula that you can derive it actually, doing this exact same process we just did, but the formula for the vertex, or the x-value of the vertex, or the axis of symmetry, is x is equal to negative b over 2a. So if we just apply this-- but, you know, this is just kind of mindless application of a formula. I wanted to show you the intuition why this formula even exists. But if you just mindlessly apply this, you'll get-- what's b here? So x is equal to negative-- b here is 8. 8 over 2 times a. a right here is a negative 2. 2 times negative 2. So what is that going to be equal to? It is negative 8 over negative 4, which is equal to 2, which is the exact same thing we got by reasoning it out. And when x is equal to 2, y is equal to 16. Same exact result there. That's the point 2 comma 16.

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We need to find the vertex and the axis of symmetry of this graph. The whole point of doing this problem is so that you understand what the vertex and axis of symmetry is. And just as a bit of a refresher, if a parabola looks like this, the vertex is the lowest point here, so this minimum point here, for an upward opening problem.

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In this video, you will be exposed to the vertex formula.  This formula is used to find the maximum or minimum point on a parabola.

Source: Susan Jones.  "How to Memorize the Vertex Formula for a Parabola."  OER Commons, 19 May, 2016.  Retrieved 8 May, 2017
CC-BY 4.0

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How to memorize the vertex formula for a parabola

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3-minute video about "Symbol Imagery" (taking a picture with your mind) and practice applying that to remember that the x-coordinate of the vertex of a parabola is -b/2a .

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The following website also shows the parts of a Parabola and lists relevant information pertaining to each part.
https://www.boundless.com/algebra/textbooks/boundless-algebra-textbook/quadratic-functions-and-factoring-6/graphs-of-quadratic-functions-44/parts-of-a-parabola-384-17430/
Source: Boundless. "Parts of a Parabola." Boundless Algebra Boundless, 14 Feb. 2017. Retrieved 8 May, 2017
CC-BY-SA 4.0

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The following image shows the graph of a quadratic (a parabola) and identifies the parts. You will learn how to find these parts in this Padlet.

Source: CK-12.  "Finding the Parts of a Parabola."  CK-12 Foundation,  2017.  Retrieved 8 May, 2017
CC-BY-NC 3.0

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Use the following website to familiarize yourself with a quadratic function.
https://www.boundless.com/algebra/textbooks/boundless-algebra-textbook/quadratic-functions-and-factoring-6/introduction-to-quadratic-functions-43/what-is-a-quadratic-function-127-10903/

Source: Boundless. "What is a Quadratic Function?." Boundless Algebra Boundless, 2 Sep. 2016. Retrieved 8 May, 2017
CC-BY-SA 4.0