Answer
Solution set: $\displaystyle \{-\frac{3}{2}\}$
Work Step by Step
We can factor in pairs here:
$x^{2}(2x+3)+(2x+3)=0$
$(2x+3)(x^{2}+1)=0$
$x^{2}+1$ has no real zeros (does not yield zero for any real x).
$2x+3=0$ when $x=-\displaystyle \frac{3}{2}$
Solution set: $\displaystyle \{-\frac{3}{2}\}$
