Solve the following equation:

$$
12 a b x^{2}-\left(16 a^{2}-9 b^{2}\right) x-12 a b=0
$$
$$
\begin{gathered}
12 a b x^{2}-\left(16 a^{2}-9 b^{2}\right) x-12 a b=4 a x(3 b x-4 a)+3 b(3 b x-4 a)= \\
=(3 b x-4 a)(4 a x+3 b)=0
\end{gathered}
$$

therefore

$$
x_{1}=\frac{4 a}{3 b} \text { and } x_{2}=-\frac{3 b}{4 a}
$$