Given:

$\left(\frac{4 x-3}{2 x+1}\right)-10\left(\frac{2 x+1}{4 x-3}\right)=3$

Putting $\frac{4 x-3}{2 x+1}=y$, we get:

$\begin{array}{l}
y-\frac{10}{y}=3 \\
\Rightarrow \frac{y^{2}-10}{y}=3 \\
\Rightarrow y^{2}-10=3 y \\
\Rightarrow y^{2}-3 y-10=0 \\
\Rightarrow y^{2}-(5-2) y-10=0 \\
\Rightarrow y^{2}-5 y+2 y-10=0 \\
\Rightarrow y(y-5)+2(y-5)=0 \\
\Rightarrow(y-5)(y+2)=0
\end{array}$
$\text { Case I: }$
$\begin{array}{l}
\Rightarrow y-5=0 \text { or } y+2=0 \\
\Rightarrow y=5 \text { or } y=-2
\end{array}$
If $y=5$, we get
$\begin{array}{l}
\frac{4 x-3}{2 x+1}=5 \\
\Rightarrow 4 x-3=5(2 x+1) \quad[\text { On cross multiplying] } \\
\Rightarrow 4 x-3=10 x+5 \\
\Rightarrow-6 x=8 \\
\Rightarrow-6 x=8 \\
\Rightarrow x=\frac{8}{6} \\
\Rightarrow x=-\frac{4}{3}
\end{array}$
Case II:
If $y=-2$, we get:
$\begin{array}{l}
\frac{4 x-3}{2 x+1}=-2 \\
\Rightarrow 4 x-3=-2(2 x+1) \\
\Rightarrow 4 x-3=-4 x-2 \\
\Rightarrow 8 x=1 \\
\Rightarrow x=\frac{1}{8}
\end{array}$

Hence, the roots of the equation are $-\frac{4}{3}$ and $\frac{1}{8}$.