Enter all the solutions to
\[ \sqrt{4x-3}+\frac{10}{\sqrt{4x-3}}=7,\]separated by commas.
We start by substituting $u=\sqrt{4x-3}$. Then it is easy to solve for $u$:
\begin{align*}
u + \frac{10}{u} &= 7 \\
u^2 + 10 &= 7u \\
u^2 - 7u + 10 &= 0 \\
(u - 5)(u - 2) &= 0
\end{align*}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 $x=\boxed{\frac 74,7}$.