Answer
{$-1,\frac{5}{2}$}
Work Step by Step
Using the rules of factoring trinomials to factor, we obtain:
$2x^{2}-3x-5=0$
$2x^{2}+2x-5x-5=0$
$2x(x+1)-5(x+1)=0$
$(x+1)(2x-5)=0$
$(x+1)=0$ or $(2x-5)=0$
$x=-1$ or $x=\frac{5}{2}$
Therefore, the solution set is {$-1,\frac{5}{2}$}.
