Chapter 10 - Quadratic Equations - 10.1 - Quadratic Equations - Problem Set 10.1 - Page 442: 8

{$-4,-12$}

Using the rules of factoring trinomials, we obtain: $x^{2}+16x+48=0$ $x^{2}+4x+12x+48=0$ $x(x+4)+12(x+4)=0$ $(x+4)(x+12)=0$ $(x+4)=0$ and $(x+12)=0$ $x=-4$ and $x=-12$ Therefore, the solution set is {$-4,-12$}.