solve for x. x^2-3x=28?

Subtract 28 from both sides, giving you x^2-3x-28=0. This factors out to (x+4)(x-7), because 4 times -7 is -28, and 7 plus -4 is -3. So x is now either (x+4) or (x-7). You write that as "x+4=0 or x-7=0", then solve for x in both equations. The answer(s) come out to be x= -4 or x= 7

x^2 - 3x = 28
x^2 - 3x - 28 = 0
x^2 + 4x - 7x - 28 = 0
(x^2 + 4x) - (7x + 28) = 0
x(x + 4) - 7(x + 4) = 0
(x + 4)(x - 7) = 0

x + 4= 0
x = -4

x - 7x = 0
x = 7

∴ x = -4 , 7

x² - 3x - 28 = 0
(x - 7) (x + 4) = 0
x = 7 , x = - 4

x^2 - 3x - 28 = 0

Quadratic equation yields:

x = 4, -7