use factors to solve 2n squared + 8n -24 =0?

n^2 + 4n - 12 = 0
Now you're looking for two numbers whose product is -12, and whose sum is +4.
(n+6) (n-2) = 0

= 2n² + 8n - 24
= 2(n² + 4n - 12)
= 2(n + 6)(n - 2)

n + 6 = 0, n = - 6
n - 2 = 0, n = 2

Answer: n = - 6, 2

Proof (where n is - 6):
2n² + 8n = 24
2(- 6²) + 8(- 6) = 24
2(36) - 48 = 24
72 - 48 = 24
24 = 24

Proof (where n is 2):
2(2²) + 8(2) = 24
2(4) + 16 = 24
8 + 16 = 24
24 = 24

n² + 4n - 12 = 0
(n + 6)(n - 2) = 0
n = - 6 , n = 2

2n^2 + 8n - 24 = 0
(2n + 12)(n - 2) = 0

This means:
(2n + 12) = 0
and
(n - 2) = 0

2n + 12 = 0
2n = -12
n = -6

OR

n -2 = 0
n = 2

2n^2 + 8n - 24 = 0
(2n - 4)(n + 6) = 0

2n - 4 = 0
2n = 4
n = 4/2
n = 2

n + 6 = 0
n = -6

∴ n = 2 , -6

2n*n+8n-24=0=
2*(n*n+4n-12)=
n*n+4n-12

factors of 12: [(12,1),(6,2),(3,4)]

since the 0th power term is neg (-12), this implies we factor into a positive and a negative.

So out of our 3 factors of 12, figure out which one has a difference of 4.

1-12=-11
2-6=-4
3-4= -1

12-1=11
6-2=4
4-3=1

Ok, so (6) and (-2).

that leaves us with

(n+6)*(n-2)=2

setting each part to 0 and solving

n+6=0
n-2=0

thus m={-6,2}

This is how you do it.
2n^2+8n-24=0
2n^2+12n-4n+24=0 (you split the 8 as 12 and -4)
2n(n+6)-4(n+6)=0
(n+6)(2n-4)=0
(n+6)2(n-2)=0
2(n+6)(n-2)=0
n+6=0
n=-6
n-2=0
n=2

Use the form;

AX2 + BX + C = 0

then;

x = (-b +/- sqrt(b^2 - 4ac))/2a

so plug and chug

x = (-8 +/- sqrt(8^2 - 4(2)(-24)) / 2(2)
x = (-8 +/- sqrt(64 + 192) / 4
x = (-8 +/- sqrt(256)) / 4
x = (-8 +/- 16) / 4
x = -24/4 & 8/4
x = -6 & 2

2n^2 +8n -24 =0
n^2 +4n -12 =0
(n-2)(n+6) =0
n-2=0, n+6 =0
n=2,-6

2x^2 + 8n - 24 = 0 factors to

(2n - 4)(n + 6) = 0

Set both factors equal to 0
2n - 4 = 0
n + 6 = 0

Solve for n
2n = 4
n = 2

n + 6 = 0
n = -6