Answer
{$-\frac{5}{3},\frac{1}{2}$}
Work Step by Step
Using the rules of factoring trinomials, we obtain:
$6y^{2}+7y-5=0$
$6y^{2}-3y+10y-5=0$
$3y(2y-1)+5(2y-1)=0$
$(2y-1)(3y+5)=0$
$(2y-1)=0$ and $(3y+5)=0$
$y=\frac{1}{2}$ and $y=-\frac{5}{3}$
Therefore, the solution set is {$-\frac{5}{3},\frac{1}{2}$}.
