#### Answer

$x=\left\{ \dfrac{2-2\sqrt{2}}{3},\dfrac{2+2\sqrt{2}}{3} \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $ (2-3x)^2=8 ,$ take the square root of both sides (Square Root Principle). Then simplify the resulting radical and use the properties of equality to isolate the variable.
$\bf{\text{Solution Details:}}$
Taking the square root of both sides (Square Root Principle), the given equation is equivalent to
\begin{array}{l}\require{cancel}
\sqrt{(2-3x)^2}=\pm\sqrt{8}
\\\\
2-3x=\pm\sqrt{4\cdot2}
\\\\
2-3x=\pm\sqrt{(2)^2\cdot2}
\\\\
2-3x=\pm2\sqrt{2}
\\\\
-3x=-2\pm2\sqrt{2} \\\\ x=\dfrac{-2\pm2\sqrt{2}}{-3} \\\\ x=\dfrac{2\pm2\sqrt{2}}{3} .\end{array}
Hence, the solutions are $
x=\left\{ \dfrac{2-2\sqrt{2}}{3},\dfrac{2+2\sqrt{2}}{3} \right\}
.$