Answer
$\dfrac{-6a}{2a+1}$
Work Step by Step
Factoring the expressions and then cancelling common factors between the numerator and the denominator, then the given expression, $
\dfrac{18a-12a^2}{4a^2+4a+1}\cdot\dfrac{4a^2+8a+3}{4a^2-9}
$, simplifies to
\begin{array}{l}\require{cancel}
\dfrac{-6a(-3+2a)}{(2a+1)(2a+1)}\cdot\dfrac{(2a+1)(2a+3)}{(2a-3)(2a+3)}
\\\\=
\dfrac{-6a(\cancel{2a-3})}{(\cancel{2a+1})(2a+1)}\cdot\dfrac{(\cancel{2a+1})(\cancel{2a+3})}{(\cancel{2a-3})(\cancel{2a+3})}
\\\\=
\dfrac{-6a}{2a+1}
.\end{array}
