## College Algebra (10th Edition)

$|x^2+x|=12$ $x^2+x=12$ 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}.