#### Answer

Solution set: $ (-\displaystyle \infty, -\frac{3}{2})\cup(0,\infty)$

#### Work Step by Step

Follow the "Procedure for Solving Polynomial lnequalities",\ p.412:
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.
$2x^{2}+3x > 0$
$f(x)=2x^{2}+3x \quad $...factor the trinomial...
$f(x)=x(2x+3)$
$x(2x+3) > 0$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$x(2x+3) = 0$
$x=0$ or $x=-\displaystyle \frac{3}{2}$
3. Locate these boundary points on a number line, thereby dividing the number line into intervals.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & (-\infty, -\frac{3}{2}) & (-\frac{3}{2},0) & (0,\infty)\\
a=test.val. & -3 & -1 & 1\\
f(a) & (-3)(-3) & (-1)(1) & (1)(5)\\
f(a) > 0 ? & T & F & T
\end{array}$
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: $ (-\displaystyle \infty, -\frac{3}{2})\cup(0,\infty)$