## Elementary Algebra

{$-5,\frac{3}{2}$}
Using the rules of factoring trinomials to factor the polynomial, we obtain: $-7n-2n^{2}=-15$ $-2n^{2}-7n+15=0$ $2n^{2}+7n-15=0$ $2n^{2}-3n+10n-15=0$ $n(2n-3)+5(2n-3)=0$ $(2n-3)(n+5)=0$ Now, we equate the two factors to zero and solve: $(2n-3)(n+5)=0$ $(2n-3)=0$ or $(n+5)=0$ $n=\frac{3}{2}$ or $n=-5$ Therefore, the solution set is {$-5,\frac{3}{2}$}.