Two square roots

Conditions. $x\ge-3$ and $x\ge1/2$, so $x\ge1/2$.

First squaring. Isolate: $\sqrt{2x-1}=4-\sqrt{x+3}$. The RHS must be $\ge0$, so $\sqrt{x+3}\le4$, i.e. $x\le13$.
Square:
$$2x-1=16-8\sqrt{x+3}+x+3,$$
i.e.
$$x-20=-8\sqrt{x+3}.$$
For the RHS to be $\ge0$, we need $x\le20$ (already have $x\le13$). Also $x-20<0$ and $-8\sqrt{x+3}\le0$, so both sides are negative; squaring is valid.

Second squaring. $(x-20)^2=64(x+3)$, i.e. $x^2-40x+400=64x+192$, i.e. $x^2-104x+208=0$.

Roots.
$$x=\frac{104\pm\sqrt{10816-832}}{2}=\frac{104\pm\sqrt{9984}}{2}.$$
$\sqrt{9984}=\sqrt{64\cdot156}=8\sqrt{156}=16\sqrt{39}$.
So $x=52\pm8\sqrt{39}$.

Check conditions. $8\sqrt{39}\approx49.96$, so $x_1\approx2.04$ (valid, $\ge1/2$ and $\le13$) and $x_2\approx101.96$ (invalid, $>13$).

Conclusion. $x=52-8\sqrt{39}$.