Solve for x: 2x(x-3) = x^2+3x-18?

2x(x - 3) = x² + 3x - 18
2x = (x² + 3x - 18)/(x - 3)
2x = x + 6
x = 6

Answer: x = 6, x = 3

Proof (x = 6):
2(6)(6 - 3) = 6² + 3(6) - 18
12(3) = 36 + 18 - 18
36 = 36

Proof (x = 3):
2(3)(3 - 3) = 3² + 3(3) - 18
6(0) = 9 + 9 - 18
0 = 0

2x(x - 3) = x^2 + 3x - 18
2x^2 - 6x = x^2 + 3x - 18
2x^2 - x^2 - 6x - 3x + 18 = 0
x^2 - 9x + 18 = 0
(x - 6)(x - 3) = 0

x - 6 = 0
x = 6

x - 3 = 0
x = 3

∴ x = 6 , 3

2x² - 6x = x² + 3x - 18
x² - 9x + 18 = 0
(x - 6)(x - 3) = 0
x = 6 , x = 3

2x(x-3) = x^2+3x-18

2x^2 - 6x - x^2 - 3x + 18 =0
x^2 - 9x + 18 =0
( x-6 ) ( x-3) =0
x-6=0
x=6

x-3=0
x=3

x=6, x=3

Distribute the 2x on the left. Get everything over to the left side. You should end up with something in the form of ax^2 + bx + c = 0. Solve the quadratic using factoring, completing the square, or the quadratic formula.