How do you simplify this square root of sum: $\sqrt{7+4\sqrt3}$?

I came around this expression when solving a problem.

$$\sqrt{7+4\sqrt{3}}$$

WolframAlpha says it equals $2+\sqrt{3}$. We can confirm it like this $$\left(2+\sqrt{3}\right)^2 \;=\; 4+4\sqrt{3} + 3 \;=\; 7 + 4\sqrt{3}.$$

However, the only way I can think of how to simplify that expression in hand is guessing. Is there a better way of calculating square root of a sum like that one?

If you want $\sqrt{7+4\sqrt{3}}=a+b\sqrt d$ (where $a$ and $b$ are rational numbers, and $d$ is a square-free integer) then $7+4\sqrt{3} = a^2+db^2+2ab\sqrt d$, which yields the natural choice $d=3$.

Then $a^2+3b^2=7$ and $2ab=4$ can be solved in the rational numbers, and you find $a=2,b=1$ (because $a^2+3 \cdot (2/a)^2=7$ has solutions $±2$ and $±\sqrt 3$).

${\sqrt {7 + 4{\sqrt3}}} = {\sqrt {7 + \sqrt{48}}}$

The latter can be solved by the formula:

$\sqrt {a + \sqrt{b} } = \sqrt{ {a + \sqrt{a^2 -b}}\over2} +\sqrt{ {a - \sqrt{a^2 -b}}\over2} $