Strategies to denest nested radicals $\sqrt{a+b\sqrt{c}}$
There do exist general denesting algorithms employing Galois theory, but for the simple case of quadratic algebraic numbers we can employ a simple rule that I discovered as a teenager.
Simple Denesting Rule $\rm \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $
Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\, b^2 $
and, $ $ furthermore, $\rm\ w\:$ has $ $ trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2a$
Here $\:61-24\sqrt{5}\:$ has norm $= 29^2.\:$ $\rm\, \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 29\ $ yields $\ 32-24\sqrt{5}\:$
and this has $\rm\ \sqrt{trace}\: =\: 8,\ \ thus,\ \ \ \color{brown}{dividing \ it \ out}\, $ of this yields the sqrt: $\,\pm( 4\,-\,3\sqrt{5}).$
See here for a simple proof of the rule, and see here for many examples of its use.
There are the following identities. $$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$ and $$\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}},$$ where all numbers under radicals are non-negatives.
For example: $$\sqrt{5+2\sqrt6}=\sqrt{5+\sqrt{24}}=\sqrt{\frac{5+\sqrt{5^2-24}}{2}}+\sqrt{\frac{5-\sqrt{5^2-24}}{2}}=\sqrt3+\sqrt2.$$ This is interesting, when $a$ and $b$ are rationals and $a^2-b$ is a square of a rational number.
The first identity is true because $$\left(\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\right)^2=$$ $$=\frac{a+\sqrt{a^2-b}}{2}+\frac{a-\sqrt{a^2-b}}{2}+2\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\cdot\sqrt{\frac{a-\sqrt{a^2-b}}{2}}=a+\sqrt{b}.$$
You can derive a formula for $\sqrt{a+b\sqrt{c}}$. You will have to assume that $\sqrt{a+b\sqrt{c}}$ can be rewritten as the sum of two surds (radicands). So $$\sqrt{a+b\sqrt{c}}=\sqrt{d}+\sqrt{e}$$
Squaring both sides yields $$a+b\sqrt{c}=d+e+2\sqrt{de}$$
From that, we can see that $a=d+e$ so $e=a-d$ and $b\sqrt{c}=2\sqrt{de}\rightarrow b^{2}c=4de$.
Substituting $e$ with $a-d$ gives $b^{2}c=4d(a-d)$. So $b^{2}c=4ad-4d^{2}$. Rearranging the terms gives us $4d^{2}-4ad+b^{2}c=0$
Using the Quadratic Equation, we have $$d=\frac {a\pm\sqrt{a^{2}-b^{2}c}}{2}$$
And since $a=d+e$, $e$ is the conjugate of $d$. So $e=\frac {a-\sqrt{a^{2}-b^{2}c}}{2}$ and $d=\frac {a+\sqrt{a^{2}-b^{2}c}}{2}.\,$ Thus
$$\sqrt{a+b\sqrt{c}}\,=\, \sqrt{\frac {a+\sqrt{a^{2}-b^{2}c}}{2}} +\sqrt{\frac {a-\sqrt{a^{2}-b^{2}c}}{2}}$$