Answer
The real solution set is {-4,3}.
Work Step by Step
$|x^2+x|=12$
$x^2+x=12$ or $x^2+x=-12$
$x^2+x-12=0$ or $x^2+x+12=0$
$(x-3)(x+4)=0$ or $x=\frac{-1+\sqrt {1-48}}{2}$ or $x=\frac{-1-\sqrt {1-48}}{2}$
$(x-3)=0$ or $(x+4)=0 $ or $x=\frac{-1+\sqrt {-47}}{2}$ or $x=\frac{-1-\sqrt {-47}}{2}$
$x=3 $ or $x=-4$ or $x=\frac{-1+i\sqrt {47}}{2}
$ or $x=\frac{-1-i\sqrt {47}}{2}$
The real solution set is {-4,3}.
