$\begin{array}{l}
3\left(\frac{3 x-1}{2 x+3}\right)-2\left(\frac{2 x+3}{3 x-1}\right)=5, x \neq \frac{1}{3},-\frac{3}{2} \\
\Rightarrow \frac{3(3 x-1)^{2}-2(2 x+3)^{2}}{(2 x+3)(3 x-1)}=5 \\
\Rightarrow \frac{3\left(9 x^{2}-6 x+1\right)-2\left(4 x^{2}+12 x+9\right)}{6 x^{2}+7 x-3}=5 \\
\Rightarrow \frac{27 x^{2}-18 x+3-8 x^{2}-24 x-18}{6 x^{2}+7 x-3}=5 \\
\Rightarrow \frac{19 x^{2}-42 x-15}{6 x^{2}+7 x-3}=5 \\
\Rightarrow 19 x^{2}-42 x-15=30 x^{2}+35 x-15 \\
\Rightarrow 11 x^{2}+77 x=0 \\
\Rightarrow 11 x(x+7)=0 \\
\Rightarrow x=0 \text { or } x+7=0 \\
\Rightarrow x=0 \text { or } x=-7
\end{array}$

Hence, 0 and $-7$ are the roots of the given equation.