Answer
{$-\frac{5}{6},2$}
Work Step by Step
$-6x^{2}+7x=-10$
Adding 10 on both sides, $-6x^{2}+7x+10=-10+10$
Simplifying, $-6x^{2}+7x+10=0$
$6x^{2}-7x-10=0$
$6x^{2}-12x+5x-10=0$
$6x(x-2)+5(x-2)=0$
$(x-2)(6x+5)=0$
$(x-2)=0$ or $(6x+5)=0$
$x=2$ or $x=-\frac{5}{6}$
Check:
Substituting $x=2$ in the equation,
$6(2)^{2}-7(2)-10=0$
$24-14-10=0$
$24-24=0$
$0=0$
Substituting $x=-\frac{5}{6}$ in the equation,
$6(-\frac{5}{6})^{2}-7(-\frac{5}{6})-10=0$
$\frac{6\times25}{36}+\frac{35}{6}-10=0$
$\frac{25}{6}+\frac{35}{6}-10=0$
$\frac{60}{6}-10=0$
$10-10=0$
$0=0$
Both values check and the solution set is {$-\frac{5}{6},2$}
