Question

$x^{2} +4x-12=0$

Answer 1

$x^{2} + 4x -12 =0 \\ \Delta=4^2-4.1.(-12) =16+48=64 \\ x=\frac{-4+-\sqrt{64}}{2.1}=\frac{-4+-8}{2} \\ x_1=-6 \\ x_2=2 \\ \boxed{S=\{-6,2 \}}$