Answer
{$-\frac{2}{5},1$}
Work Step by Step
$5x^{2}-3x-2=0$
$5x^{2}-5x+2x-2=0$
$5x(x-1)+2(x-1)=0$
$(x-1)(5x+2)=0$
$(x-1)=0$ or $(5x+2)=0$
$x=1$ or $x=-\frac{2}{5}$
Check:
Substituting $x=1$ in the equation,
$5(1)^{2}-3(1)-2=0$
$5-3-2=0$
$5-5=0$
$0=0$
Substituting $x=-\frac{2}{5}$ in the equation,
$5(-\frac{2}{5})^{2}-3(-\frac{2}{5})-2=0$
$\frac{20}{25}+\frac{6}{5}-2=0$
$\frac{4}{5}+\frac{6}{5}-2=0$
$\frac{10}{5}-2=0$
$2-2=0$
$0=0$
Both values check and the solution set is {$-\frac{2}{5},1$}
