Answer
$[-\frac{2}{3},\frac{3}{2}]$ or $\{x|-\frac{2}{3}\le x\le\frac{3}{2}\}$.
Work Step by Step
Step 1. $6x^2\le6+5x \longrightarrow 6x^2-5x-6\le0 \longrightarrow (2x-3)(3x+2)\le0$.
Step 2. Identify boundary points $x=-\frac{2}{3},\frac{3}{2}$ and form intervals $(-\infty,-\frac{2}{3}],[-\frac{2}{3},\frac{3}{2}],[\frac{3}{2},\infty)$.
Step 3. Choose test values $x=-1,0,2$ to test the inequality and get results $False,\ True,\ False$.
Step 4. Thus we have the solution $[-\frac{2}{3},\frac{3}{2}]$ or $\{x|-\frac{2}{3}\le x\le\frac{3}{2}\}$.
