Answer
{$-\frac{2}{5},\frac{3}{4}$}
Work Step by Step
Factoring the equation and setting the factors equal to zero, we obtain:
$20y^{2}-7y-6=0$
$20y^{2}+8y-15y-6=0$
$4y(5y+2)-3(5y+2)=0$
$(5y+2)(4y-3)=0$
$(5y+2)=0$ or $(4y-3)=0$
$y=-\frac{2}{5}$ or $y=\frac{3}{4}$
Therefore, the solution set is {$-\frac{2}{5},\frac{3}{4}$}.
