Chapter 10 - Quadratic Equations - Chapter 10 Review Problem Set - Page 467: 14

{$-3,9$}

Using the rules for the factoring of trinomials, we obtain: $x(x-6)=27$ $x^{2}-6x-27=0$ $x^{2}+3x-9x-27=0$ $x(x+3)-9(x+3)=0$ $(x+3)(x-9)=0$ $(x+3)=0$ and $(x-9)=0$ $x=-3$ and $x=9$ Therefore, the solution set is {$-3,9$}.