Answer
{$-\frac{3}{4},1$}
Work Step by Step
$-4x^{2}+x=-3$
Adding 3 on both sides, $-4x^{2}+x+3=-3+3$
Simplifying, $-4x^{2}+x+3=0$
$4x^{2}-x-3=0$
$4x^{2}-4x+3x-3=0$
$4x(x-1)+3(x-1)=0$
$(x-1)(4x+3)=0$
$(x-1)=0$ or $(4x+3)=0$
$x=1$ or $x=-\frac{3}{4}$
Check:
Substituting $x=1$ in the equation,
$4(1)^{2}-(1)-3=0$
$4-1-3=0$
$4-4=0$
$0=0$
Substituting $x=-\frac{3}{4}$ in the equation,
$4(-\frac{3}{4})^{2}-(-\frac{3}{4})-3=0$
$\frac{36}{16}+\frac{3}{4}-3=0$
$\frac{9}{4}+\frac{3}{4}-3=0$
$\frac{12}{4}-3=0$
$3-3=0$
$0=0$
Both values check and the solution set is {$-\frac{3}{4},1$}
