Answer
$[-\frac{1}{2},3]$ or $\{x|-\frac{1}{2}\le x\le3\}$.
Work Step by Step
Step 1. $2x^2\le5x+3 \longrightarrow 2x^2-5x-3\le0 \longrightarrow (2x+1)(x-3)\le0$.
Step 2. Identify boundary points $x=-\frac{1}{2},3$ and form intervals $(-\infty,-\frac{1}{2}],[-\frac{1}{2},3],[3,\infty)$.
Step 3. Choose test values $x=-1,0,4$ to test the inequality and get results $False,\ True,\ False$.
Step 4. Thus we have the solution $[-\frac{1}{2},3]$ or $\{x|-\frac{1}{2}\le x\le3\}$.
