Answer
{$-4,\frac{7}{3}$}
Work Step by Step
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 {$-4,\frac{7}{3}$}.
