Answer
$x$ can be any real number except $3$; that is,$$x\in \mathbb{R} -\{ 3 \}$$
Work Step by Step
Please note that for any real number $y$, we have $y^2 \ge 0$ and the equality holds if and only if $y=0$.
Hence, the inequality $(x-3)^2 > 0$ holds for any real number $x$ except the one satisfying $x-3=0$, $x=0$. Thus,$$x\in \mathbb{R} -\{ 3 \}$$
