Answer
$(-\infty,-1]\cup (\frac{5}{2},3]$
Work Step by Step
Rewrite the inequality as $x-\frac{6-x}{2x-5}\leq0$
which gives $\frac{(x+1)(x-3)}{2x-5}\leq0$
Test signs in intervals $(-\infty,-1), (-1,5/2), (5/2,3), (3,\infty)$, we get $-,+,-,+$ for the function.
Test end points and we get the solution as $x\in(-\infty,-1]\cup (\frac{5}{2},3]$