Answer
$(-\infty,2)\cup(2,\infty)$
Work Step by Step
Step 1. The term $(x-2)^2$ is a perfect square and it will always be positive or zero.
Step 2. Because the above term is a denominator, we need to exclude $x=2$
Step 3. We can conclude that the solutions are all real numbers except $x=2$, or in interval notation, $(-\infty,2)\cup(2,\infty)$
