Answer
$(-\infty, -1]\cup[\frac{3}{2},\infty)$
Work Step by Step
Step 1. Move all the term to the left side and factor, we have $2x^2-x-3\geq0\longrightarrow (x+1)(2x-3)\geq0$
Step 2. Identify the boundary points as $x=-1$ and $x=\frac{3}{2}$
Step 3. The boundary points separate the number line into three intervals $(-\infty, -1)$, $(-1, \frac{3}{2})$ and $(\frac{3}{2},\infty)$
Step 4. Use test points $x=-2,0,2$ for each interval, we can determine the signs of the left side of the inequality as $+, - , +$
Step 5. Based on the signs and consider the boundary points, we can write the solution set as $(-\infty, -1]\cup[\frac{3}{2},\infty)$
