Technique for simplifying, e.g. $\sqrt{ 8 - 4\sqrt{3}}$ to $\sqrt{6} - \sqrt{2}$ How to find the square root of an irrational expression, to simplify that root. e.g.:
$$
\sqrt{ 8 - 4\sqrt{3} } = \sqrt{6} - \sqrt{2}
$$
Easy to verify:
\begin{align}
(\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} +2 
 = 8 - 4 \sqrt{3}
\end{align}
But how to work it out in the first place? I feel there's a standard technique (Completing-the-square? Quadratic formula?), but don't recall it or what it's called...
BTW: this came up in verifying equivalence of different calculations of $\cos{75°}$ (the above divided by $4$), as $\cos{\frac{90°+60°}{2}}$ vs $\cos{(45°+30°)}$, from 3Blue1Brown's lockdown video on complex numbers and trigonometry.

I'm assuming you're looking for a formula to de-nest $\sqrt{a+b\sqrt{c}}$.
To derive such a formula, equate this to $x+y\sqrt{c}$.
$a+b\sqrt{c}=(x+y\sqrt{c})^2=x^2+cy^2+2xy\sqrt{c}$
$a+b\sqrt{c}=x^2+cy^2+2xy\sqrt{c}$
On matching up coefficients, we get:
$
\begin{cases}
x^2+cy^2=a\\
2xy=b
\end{cases}
$
Substitute $y=\frac{b}{2x}$:
$\rightarrow x^2+c(\frac{b^2}{4x^2})=a$
$u:=x^2\rightarrow u+\frac{cb^2}{4u}=a$
$\rightarrow u^2-au+\frac{b^2c}{4}=0$
$\rightarrow u=\frac{a\pm\sqrt{a^2-b^2c}}{2}$
$\because u=x^2\rightarrow x=\sqrt{\frac{a\pm\sqrt{a^2-b^2c}}{2}}$
Now solve for $y$:
$y=\frac{b}{2}(\sqrt{\frac{2}{a\pm\sqrt{a^2-b^2c}}})$
And you're done! This tells us that:
$$\sqrt{a+b\sqrt{c}}=\sqrt{\frac{a+\sqrt{a^2-b^2c}}{2}}+(\frac{b}{2}(\sqrt{\frac{2}{a+\sqrt{a^2-b^2c}}}))\sqrt{c}$$
