How do you solve 3x²-8x=x²-8x+72?

3x^2 - x^2 - 8x + 8x - 72 = 0
2x^2 - 72 = 0
x^2 - 36 = 0
(x - 6)(x + 6) =0
x = 6 x = - 6

2x² - 72 = 0
x² - 36 = 0
x = ± 6

rearrange your equation:

3x^2-x^2-8x+8x=72

this leaves us with:

2x^2=72

devide both sides with 2:

x^2=36

get the squareroot of both sides:
the value of are:

x=6 and x=-6

3x^2-8x-x^2+8x-72=0
2x^2-72=0
2(x^2-36)=0
x^2-36=0
(x+6)(x-6)=0
x= +6 , -6

3x²-8x=x²-8x+72
-x² from both sides
2x²-8x=-8x+72
+8x to both sides
2x²=72
/ both side by 2
x²=36
square root both sides
x=6,-6

3x^2 - 8x = x^2 - 8x +72

Rearrange so that you get all the x squared, x and pure numbers on one side...

2x^2 - 72 = 0

The 8x's cancelled out and you can divide the whole thing by 2 since is a common factor...

x^2 - 36 = 0

That looks like a difference of two squares...

(x + 6)(x - 6) = 0

So x = 6 and -6

3x^2 - 8x = x^2 - 8x + 72

Add 8x to each side:

3x^2 = x^2 + 72, so

2x^2 = 72, so

x^2 = 36, so

x = +/- 6.

3x^2-8x= x^2-8x+72. Bring everything to one side to get:

2x^2-72=0, or x^2-36=0. Thus x=6 or x=-6

3x^2 - 8x = x^2 - 8x + 72
3x^2 - x^2 - 8x + 8x - 72 = 0
2x^2 - 72 = 0
x^2 - 36 = 0
(x + 6)(x - 6) = 0

x + 6 = 0
x = -6

x - 6 = 0
x = 6

∴ x = -6 , 6