Answer
$( -\infty, -1] \cup \left[ \frac{3}{2}, \infty\right)$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a polynomial function.
$2x^2\geq x+3$,
$2x^2-x-3 \geq0$,
$2x^2+2x-3x-3 \geq 0$,
$2x(x+1)-3(x+1) \geq 0,$
$(2x-3)(x+1)\geq0,$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$(2x-3)(x+1)=0$
$x=-1$ or $x=\frac{3}{2}$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \geq 0 ? \\
& &(2a-3)(a+1)& \\
(-\infty, -1) & -2 & (-)(-)=(+) & T\\
(-1, \frac{3}{2}) & 0 & (-)(+)=(-) & F\\
(\frac{3}{2},\infty) & 2 & (+)(+)=(+) & T
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $( -\infty, -1] \cup \left[ \frac{3}{2}, \infty\right)$
