jmc

Subtracting $2$ from both sides, we get $$\begin{array}{rl}\frac{2x-5}{x+3}-2& \ge 0\\ \frac{2x-5-2(x+3)}{x+3}& \ge 0\\ \frac{-11}{x+3}& \ge 0.\end{array}$$Therefore, we must have $x+3<0,$ so $x<-3.$ Therefore, the solution set is $\boxed{(-\infty ,-3)}.$