How do you solve x^2+x-8=4?

1 Answer
Feb 10, 2016

x^2 + x - 8 - 4 = 0
x^2 + x - 12 = 0
x^2 +4 x - 3x -12 = 0
x(x+4) - 3(x+4) =0
(x+4)*(x-3) =0
x+4=0, x-3=0
x=-4, x=3