jmc

We start by substituting $u=\sqrt{4x-3}$. Then it is easy to solve for $u$: $$\begin{array}{rl}u+\frac{10}{u}& =7\\ u^2+10& =7u\\ u^2-7u+10& =0\\ (u-5)(u-2)& =0\end{array}$$Thus, we must have $u=2$ or $u=5$.

If $u=2$, we get $\sqrt{4x-3}=2$, so $4x-3=4$ and $x=\frac{7}{4}$.

If $u=5$, we get $\sqrt{4x-3}=5$ and so $4x-3=25$, yielding $x=7$.

Thus our two solutions are $x=\boxed{\frac{7}{4},7}$.