Answer
{$-4, \frac{2}{3}$}
Work Step by Step
Using the rules of factoring trinomials to factor the polynomial, we obtain:
$8-10x-3x^{2}=0$
$-3x^{2}-10x+8=0$
$3x^{2}+10x-8=0$
$3x^{2}-2x+12x-8=0$
$x(3x-2)+4(3x-2)=0$
$(3x-2)(x+4)=0$
Now, we equate the two factors to zero and solve:
$(3x-2)(x+4)=0$
$(3x-2)=0$ or $(x+4)=0$
$x=\frac{2}{3}$ or $x=-4$
Therefore, the solution set is {$-4, \frac{2}{3}$}.
