#### Answer

$(-\infty,-3)\cup[8,\infty)$

#### Work Step by Step

Step 1. Rewrite the inequality as $\frac{x+2}{x-3}-2\leq0$, $\frac{x+2-2x+6}{x-3}\leq0$, $\frac{-x+8}{x-3}\leq0$, and $\frac{x-8}{x-3}\geq0$; then graph the function $f(x)=\frac{x-8}{x-3}$ as shown in the figure.
Step 2. The zeros of the function can be identified as $x=8$
Step 3. We can identify the regions where $f(x)\geq0$ as $(-\infty,-3)\cup[8,\infty)$