#### Answer

$\color{blue}{\left\{4-i\sqrt5, 4+i\sqrt5\right\}}$.

#### 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:
$\sqrt{(x-4)^2}=\pm \sqrt{-5}
\\x-4 =\pm \sqrt{-1(5)}$
Since $\sqrt{-1}=i$, then the expression above is equivalent to:
$x-4=\pm i\sqrt{5}$
Add $4$ to both sides:
$x =4 \pm i\sqrt{5}$
Thus, the solution set is $\color{blue}{\left\{4-i\sqrt5, 4+i\sqrt5\right\}}$.