#### Answer

$\color{blue}{\left\{\dfrac{5-2i\sqrt2}{2}, \dfrac{5+2i\sqrt2}{2}\right\}}$.

#### Work Step by Step

RECALL:
If $x^2=a$, then taking the square root of both sides gives $x = \pm \sqrt{a}$.
Take the square root of both sides of the given equation to obtain:
$\sqrt{(-2x+5)^2}=\pm \sqrt{-8}
\\-2x+5 =\pm \sqrt{4(-1)(2)}
\\-2x+5 =\pm \sqrt{2^2(-1)(2)}
\\-2x+5 =\pm 2\sqrt{(-1)(2)}$
Since $\sqrt{-1}=i$, then the expression above is equivalent to:
$-2x+5=\pm 2i\sqrt{2}$
Add $-5$ to both sides:
$-2x =-5 \pm 2i\sqrt{2}$
Divide $-2$ to both sides:
$x=\dfrac{-5\pm 2i\sqrt2}{-2}
\\x=\dfrac{5\mp 2i\sqrt2}{2}$
Thus, the solution set is $\color{blue}{\left\{\dfrac{5-2i\sqrt2}{2}, \dfrac{5+2i\sqrt2}{2}\right\}}$.