Answer
$\left\{-\dfrac{1}{3},\dfrac{1}{6}\right\}$
Work Step by Step
Let $z=
(3x-1)
$. Then the given equation, $
2(3x-1)^2+5(3x-1)=-2
,$ is equivalent to
\begin{align*}
2z^2+5z&=-2
\\
2z^2+5z+2&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z+2)(2z+1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z+2=0 & 2z+1=0
\\
z=-2 & 2z=-1
\\\\
& z=-\dfrac{1}{2}
.\end{array}
Since $z=
(3x-1)
$, by back-substitution,
\begin{array}{l|r}
3x-1=-2 & 3x-1=-\dfrac{1}{2}
.\end{array}
Using the properties of equality to solve for the variable, then
\begin{array}{l|r}
3x=-2+1 & 3x=-\dfrac{1}{2}+1
\\\\
3x=-1 & 3x=\dfrac{1}{2}
\\\\
x=-\dfrac{1}{3} & x=\dfrac{\frac{1}{2}}{3}
\\\\
& x=\dfrac{1}{6}
.\end{array}
Hence, the solution set of the equation $
2(3x-1)^2+5(3x-1)=-2
$ is $\left\{-\dfrac{1}{3},\dfrac{1}{6}\right\}$.
