Answer
$\left\{-3,-2,1,2\right\}$
Work Step by Step
Let $z=
x^2+x
$. Then the given equation, $
(x^2+x)^2+12=8(x^2+x)
,$ is equivalent to
\begin{align*}
z^2+12&=8z
\\
z^2-8z+12&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z-6)(z-2)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z-6=0 & z-2=0
\\
z=6 & z=2
.\end{array}
Since $z=
x^2+x
$, by back-substitution,
\begin{array}{l|r}
x^2+x=6 & x^2+x=2
\\
x^2+x-6=0 & x^2+x-2=0
.\end{array}
Using factoring of trinomials, the equations above are equivalent to
\begin{array}{l|r}
(x+3)(x-2)=0 & (x+2)(x-1)=0
.\end{array}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|c|c|r}
x+3=0 & x-2=0 & x+2=0 & x-1=0
\\
x=-3 & x=2 & x=-2 & x=1
.\end{array}
Hence, the solution set of the equation $
(x^2+x)^2+12=8(x^2+x)
$ is $\left\{-3,-2,1,2\right\}$.
