Chapter 9 - Roots and Radicals - Chapters 1-9 Cumulative Review Problem Set - Page 433: 77

{$-6,2$}

We will factor this trinomial by breaking up the middle term of the trinomial. This is in accordance with the rules of factoring trinomials. $x^{2}+4x-12=0$ $x^{2}-2x+6x-12=0$ $x(x-2)+6(x-2)=0$ $(x-2)(x+6)=0$ $(x-2)=0$ and $(x+6)=0$ $x=2$ and $x=-6$ Therefore, the solution set is {$-6,2$}.