# data[].attributes.body
x^2

# data[].attributes.body
If you subtract 16, you get the equation x^2=-16. When you try to square both sides, you would get x= radical -4. If you square a negative, it's imaginary because negatives cannot be squared

# data[].attributes.body
This equation would have 2 imaginary solutions because when solved by using the quadratic equation inside the radical the discriminant would turn out to be - number, which results in 2 imaginary solutions.

# data[].attributes.body
The equation could not be solved by factoring because there are no two numbers that would multiply to 18 and add to 8. This could be solved by using the quadratic formula.

# data[].attributes.body
This equation would have two real solutions because once it is factored by getting (2x+4) or (x+2) and (2x+1) and making it equal to 0 there would be two solutions.

# data[].attributes.body
This equation has one solution because b^2 would equal to 400 and subtracted by -4(10)(10) would result in a 0. The answer would then simplify to one solution.

# data[].attributes.body
The equation can't be factored but it has 2 real solutions

# data[].attributes.body
when you solve this using the quadratic formula the discriminant is negative so the solution is an imaginary number

# data[].attributes.body
There is only one answer because when you factor the factors are both 3 .

# data[].attributes.body
This equation has exactly one solution because it evenly factors into a perfect square, (x+3)^2.

# data[].attributes.body
This equation has two real solutions but cannot be factored into fractions or whole numbers.

# data[].attributes.body
if the numbers are plugged into the quadratic formula, then the the number that is inside the radical becomes a negative number.

# data[].attributes.body
x would equal

# data[].attributes.body
y=1
There is only one answer because when you get the two factors, they are the same. This is because there can only be so many factors for the number 2

# data[].attributes.body
this equation does not factor because nothing multiplies to 8 and adds to 3. You can find the solutions using the quadratic formula

# data[].attributes.body
This equation has two real solutions that can be found by factoring because the 16 factors into the 10 evenly, ending up with (x+8) and (x+2). By setting both of the equations equal to zero, there are 2 real solutions.

# data[].attributes.body
you can only solve this by using the quadratic formula b/c it cannot be factored.

# data[].attributes.body
3x^2+6x+5=0
The quadratic equation can't be solved by factoring since 15 can't factor into 6. Only formula to use is the quadratic formula.

# data[].attributes.body
This equation factors out to (x-10)(x-2). By setting these equations equal to 0, you get the solutions 10 and 2, which are both real numbers, so there are two solutions.

# data[].attributes.body
Trying to factor this equation by factoring would be impossible. Factoring requires you to have whole numbers or fractions as answers. However, if solved using the quadratic formula, you can find the real solutions.

# data[].attributes.body
the discriminant ends up being negative, making the solution an imaginary number

# data[].attributes.body
x= -1 because (x+1)(x+1), the only solution to the problem is -1

# data[].attributes.body
x^2 + 4x + 20
the rad will be negative

# data[].attributes.body
x^2 +6x + 9

# data[].attributes.body
can only have imaginary answers because the radical will be negative.

# data[].attributes.body
when the 18 is subtracted to both sides it amkes the other side -16 there fore becomes an imaginary number

# data[].attributes.body
y=x^2-5x-36
Get solutions of +9 and -4

# data[].attributes.body
y=x^+4x-4
b factors into 2 and then changes to x+2
-evan fredericks

# data[].attributes.body
this equation could not be found by factoring because there is no number that multiplies to 6 and adds to 12.

# data[].attributes.body
y=x^2 4x+4

# data[].attributes.body
there would be 2 solutions because it would end up being (x+5)(x+4).

# data[].attributes.body
y=x^
-evan fredericks

# data[].attributes.body
when plugged into the quadratic formula it makes the number inside the radical a negative

# data[].attributes.body
can only be solved with quadratic formula

# data[].attributes.body
the 4 is added to both side and then squared, it ends up being x=+-

# data[].attributes.body
When solved, the solution is 1.

# data[].attributes.body
solve by using the quadratic formila

# data[].attributes.body
Subtracting 9 would get x^2=-9 and then square both sides and it would be squaring a negative

# data[].attributes.body
When factored the x will become -6 and 6. This doesn't multiply to 36. When placed in a radical it will have an imaginary # (or a negative)

# data[].attributes.body
It can only be factored to end up with one solution because it factors to X+1 and x-1

# data[].attributes.body
there will be a negative number inside of
the radical making it an imaginary number

# data[].attributes.body
the solution is -4 b/c when it factors you get a perfect square. (x+4)^2

# data[].attributes.body
since you cant factor this you use the quadratic formula to solve for the x values

# data[].attributes.body
This equation can't be factored because 3 doesn't factor into 10. You can solve by using the quadratic formula

# data[].attributes.body
there is a zero in the radical so it would solve out perfectly with only one solution. It is also a perfect square

# data[].attributes.body
Once you solve it through the use of the quadratic formula, you end up with the solutions of x=1 +3i and x= 1-3i

# data[].attributes.body
x=-2 and when you solve it becomes -40=0

# data[].attributes.body
You can factor this to make (x+5)(x+2) which means the solutions are -5 and -2, which are both real numbers

# data[].attributes.body
This can't be factored. You can factor this be plugging the a,b, and c values into the quadratic formula to solve without factoring.

# data[].attributes.body
This can not be factored but can b solved by the quadratic equation

# data[].attributes.body
This has only one solution because when factored, it's a perfect square (x+3)^2. Therefore, the quadratic should have only one solution.

# data[].attributes.body
when factored the solutions are -4,-2

# data[].attributes.body
y=1
This answer was only 1 because
-Imani moreen

# data[].attributes.body
this becomes 4x^2=-9, and then you need to go radical, which gets imaginary numbers

# data[].attributes.body
the 10 doesnt factor in 5, but it can still be solved by using the quadratic formula.

# data[].attributes.body
When put into the quadratic function it makes the number inside of the radical negative.

# data[].attributes.body
3x^2+5=41 Since this cannot be factored, you can solve by square roots to get the solution of 2 rad 3 and -2 rad 3

# data[].attributes.body
if you try to factor this it won't so you would use the quadratic formula.

# data[].attributes.body
Radical zero doesn't need plus or minus in front and factoring would get x=5

# data[].attributes.body
when the 2 is subtracted to both sides it equals x=1 which is one solution

# data[].attributes.body
When you factor the equation, you end up with the factors (x-4) and (x-3). This is then simplified down to two real solutions.

# data[].attributes.body
the negative sign under the square root becomes an imaginary number
x=-9i x=9i

# data[].attributes.body
it equals -23

# data[].attributes.body
it can factor evenly to (x+4) and (x+2). Set both of these equations equal to zero and there are 2 real numbers as solutions

# data[].attributes.body
x=1 because 1+1+2

# data[].attributes.body
This is way too difficult to factor, and no simple integers add to 27 and multiply to 30. It would be better to plug them in to the Quadratic formula.

# data[].attributes.attachment_link.description
See all the wonderful things Madelyn Aguilera has made

# data[].attributes.body
Using the quadratic formula, you can solve to find that the solution is 1

# data[].attributes.body
This would have imaginary solutions because the number in the square root is negative and greater than the numbers out of it, which would make x negative.

# data[].attributes.body
When solved using the quadratic formula, the discriminant becomes a negative, and the whole equation/fraction cannot be simplified.

# data[].attributes.body
when factored the solutions are -3,-5

# data[].attributes.body
x^2 + 6x __+9__ =  -16
(x + 3)^2 = -16
x + 3 = +/- 16
rad both sides and the answers will be 13 and -19

# data[].attributes.body
When factored, the two solutions would be -9 and 4 because they add up to -5 and multiply to -36

# data[].attributes.body
this factors to 1 and 1 so you would do x+1=0 and it's so -1 is the solution.

# data[].attributes.body
when factored x=9 x=10 because (x-10)(x-9)

# data[].attributes.body
The x and the 100 cannot be factor because it equals out to itself.

# data[].attributes.body
There is no number that multiply to 2 & add to -4 so you would need to use the quadratic formula for it.

# data[].attributes.body
x=2 because 4-4=0

# data[].attributes.body
You can factor to end up with (x+10) and (x+2) which comes to two solutions.

# data[].attributes.body
y=x^+4x+9
the 9 does not factor into 4 but it can still be solved by the quadratic function.

# data[].attributes.body
This only has 1 solution because it comes out to (x+3)^2

# data[].attributes.body
the equation becomes
x^2-2x+2x-4 which factors down to (x+2)(x-2). The answers are +2 and -2

# data[].attributes.body
x^2+3x+4=32

There is only one answer.

# data[].attributes.body
The factors are 2 and 3 because 2 and 3 multiply to 6 and add to 5