#### Answer

The solutions are $t=-\frac{2}{3}$ and $t=-\frac{4}{3}$.

#### Work Step by Step

Multiply $(3t-2)^2$ to both sides of the equation to obtain:
$(3t-2)^2\left(1+\dfrac{2}{3t-2}\right)=\dfrac{8}{(3t-2)^2} \cdot (3t-2)^2
\\(3t-2)^2+(3t-2)(2)=8
\\(3t-2)^2+2(3t-2)-8=0$
Let
$u=3t-2$
Rewrite the equation above using the variable $u$ to obtain:
$u^2+2u-8=0$
Factor the trinomial to obtain:
$(u+4)(u-2)=0$
Use the Zero-Factor Property by equating each unique factor to zero. then solve each equation:
\begin{array}{ccc}
&u+4 = 0 &\text{ or } &u-2=0
\\&u=-4 &\text{ or } &u=2
\end{array}
Since $u=3t-2$, then
\begin{array}{ccc}
&u= -4 &\text{or} &u=2
\\&3t-2=-4 &\text{or} &3t-2=2
\\&3t=-4+2 &\text{or} &3t=2+2
\\&3t=-2 &\text{or} &3t=4
\\&t=-\frac{2}{3} &\text{or} &t=\frac{4}{3}
\end{array}
Thus, the solutions are $t=-\frac{2}{3}$ and $t=-\frac{4}{3}$.