#### Answer

$x=4$.
Please refer to the step-by-step part below for the explanation.

#### Work Step by Step

$(x-4)^2\leq0$, since the square of a real number can only be non-negative, the only solution is when $(x-4)^2=0$.
Solve this equation to obtain:
\begin{align*}(x-4)^2&=0\\
\sqrt{(x-4)^2}&=\pm \sqrt{0}\\
x-4&=0\\
x&=4\\
\end{align*}
Thus, the only solution to the given inequality is $x=4$.