Answer :
x2+2x-8 =0
x2 + 4x - 2x - 8 = 0
x(x+4) - 2(x+4) = 0
(x-2)(x+4) = 0
let x- 2 = 0
x = 2
let x+4 = 0
x= -4
x2 + 4x - 2x - 8 = 0
x(x+4) - 2(x+4) = 0
(x-2)(x+4) = 0
let x- 2 = 0
x = 2
let x+4 = 0
x= -4
[tex] x^{2} +2x=8 \ add \ 1 \ to \ both \ sides\\
\\
x^{2} +x+8+1 \ in \ the \ left \ side \ you \ find \ a perfect \ square \ trinomial\\
\\
(x+1)^2=9\\
\\
So:\\
\\
x+1=\pm \sqrt{9}\\
\\
x+1=\pm 3\\
\\
x+1=3 \rightarrow x=2\\
\\
x+1=-3 \rightarrow \x=-4[/tex]
