Answer
$\left\{-\dfrac{16}{3}, -2\right\}$
Work Step by Step
Let $z=
(m+4)
$. Then the given equation, $
3(m+4)^2-8=2(m+4)
,$ is equivalent to
\begin{align*}
3z^2-8&=2z
\\
3z^2-2z-8&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z-2)(3z+4)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z-2=0 & 3z+4=0
\\
z=2 & 3z=-4
\\\\
& z=-\dfrac{4}{3}
.\end{array}
Since $z=
(m+4)
,$ by back substitution, then
\begin{array}{l|r}
m+4=2 & m+4=-\dfrac{4}{3}
\\\\
m=2-4 & m=-\dfrac{4}{3}-4
\\\\
m=-2 & m=-\dfrac{4}{3}-\dfrac{12}{3}
\\\\
& m=-\dfrac{16}{3}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }m=-2: & \text{If }m=-\dfrac{16}{3}
\\\\
3(-2+4)^2-8\overset{?}=2(-2+4) &
3\left(-\dfrac{16}{3}+4\right)^2-8\overset{?}=2\left(-\dfrac{16}{3}+4\right)
\\\\
3(2)^2-8\overset{?}=2(2) &
3\left(-\dfrac{16}{3}+\dfrac{12}{3}\right)^2-8\overset{?}=2\left(-\dfrac{16}{3}+\dfrac{12}{3}\right)
\\\\
3(4)-8\overset{?}=4 &
3\left(-\dfrac{4}{3}\right)^2-8\overset{?}=2\left(-\dfrac{4}{3}\right)
\\\\
12-8\overset{?}=4 &
3\left(\dfrac{16}{9}\right)-8\overset{?}=-\dfrac{8}{3}
\\\\
4\overset{\checkmark}=4 &
\dfrac{16}{3}-8\overset{?}=-\dfrac{8}{3}
\\\\
&
\dfrac{16}{3}-\dfrac{24}{3}\overset{?}=-\dfrac{8}{3}
\\\\
&
-\dfrac{8}{3}\overset{\checkmark}=-\dfrac{8}{3}
.\end{array}
Since all solutions satisfy the original equation, then the solution set of the equation $
3(m+4)^2-8=2(m+4)
$ is $\left\{-\dfrac{16}{3}, -2\right\}$.
