Answer
$x=-4$ and $x=3$
Work Step by Step
The given equation is
$x^2+x=12$
Subtract $12$ from each side.
$x^2+x-12=12-12$
Simplify.
$x^2+x-12=0$
Graph the related function.
$f(x)=x^2+x-12$
The $x-$intercepts are $-4$ and $3$.
Check $x=-4$.
$\Rightarrow x^2+x-12=0$
$\Rightarrow (-4)^2+(-4)-12=0$
$\Rightarrow 16-4-12=0$
$\Rightarrow 16-16=0$
$\Rightarrow 0=0$
True.
Check $x=3$.
$\Rightarrow x^2+x-12=0$
$\Rightarrow (3)^2+(3)-12=0$
$\Rightarrow 9+3-12=0$
$\Rightarrow 12-12=0$
$\Rightarrow 0=0$
True.
Hence, the solutions are $x=-4$ and $x=3$.
