Answer
$\frac{1}{2}$; $x\ne-\frac{3}{2}, 5$
Work Step by Step
Before we multiply two fractions, we should factor each to see if we can cancel any terms. Here, we can pull out common numeric factors from many of the terms.
$\frac{3(2x+3)}{3(x-5)}\times \frac{x-5}{2(2x+3)}$
Note that we can cross cancel both the $x-5$ and $2x+3$ terms. Also, we can cancel the $3$ that is in the numerator and denominator of the right-hand fraction. This leaves us with everything canceled in the numerator, which means there is an implied $1$. Since there is only $2$ left in the denominator, our final answer is $\frac{1}{2}$.
Also note that this is true only if $x\ne-\frac{3}{2}, 5$, since they make the original denominators equal to 0.
