## Elementary Algebra

{$-4,\frac{7}{3}$}
Using the rules of factoring trinomials, we obtain: $3x^{2}+5x-28=0$ $3x^{2}-7x+12x-28=0$ $x(3x-7)+4(3x-7)=0$ $(3x-7)(x+4)=0$ $(3x-7)=0$ and $(x+4)=0$ $x=\frac{7}{3}$ and $x=-4$ Therefore, the solution set is {$-4,\frac{7}{3}$}.