Answer
{$-4,\frac{5}{2}$}
Work Step by Step
Using the rules of factoring trinomials to factor the polynomial, we obtain:
$2x^{2}+3x-20=0$
$2x^{2}-5x+8x-20=0$
$x(2x-5)+4(2x-5)=0$
$(2x-5)(x+4)=0$
Now, we equate the two factors to zero and solve:
$(2x-5)(x+4)=0$
$(2x-5)=0$ or $(x+4)=0$
$x=\frac{5}{2}$ or $x=-4$
Therefore, the solution set is {$-4,\frac{5}{2}$}.
