## Elementary Algebra

Solving for the answer by factoring and then setting the factors equal to zero, we obtain: $x^{2}-8x-48=0$ $x^{2}+4x-12x-48=0$ $x(x+4)-12(x+4)=0$ $(x+4)(x-12)=0$ $x+4=0$ and $(x-12)=0$ $x=-4$ and $x=12$ Therefore, the correct solution set is {$-4,12$} instead of {$-12,4$}.