Answer
$\{-4,3\}$
Work Step by Step
Recall that $|a|=b \implies a=b$ or $a=-b$.
Thus, using the rule above yields:
$x^2+x=12\quad $ or $\quad x^2+x=-12$
Solve each equation.
For $x^2+x=12$:
Subtract $12$ from both sides then simplify.
$\begin{align*}
x^2+x-12&=12-12\\
x^2+x-12&=0\end{align*}$
Rewrite $x$ as $4x-3x$.
$x^2+4x-3x-12=0$
Group the terms.
$(x^2+4x)+(-3x-12)=0$
Factor out the GCF in each group.
$x(x+4)-3(x+4)=0$
Factor out $(x+4)$.
$(x+4)(x-3)=0$
Use Zero-Product Property, which states that if $ab=0$, then $a=0$ or $b-0$ or both are $0$ to obtain:
$x+4=0\quad$ or $\quad x-3=0$
$x=-4\quad $ or $\quad x=3$
For $x^2+x=-12$:
Add $12$ to both sides then simplify.
$\begin{align*}
x^2+x+12&=-12+12\\
x^2+x+12&=0\end{align*}$
The discriminant $\left(b^2-4ac\right)$ of the quadratic trinomial above is $-47$ so the equation has no real number solution.
Hence, the solution set of the given equation is $\{-4,3\}$.
