Simplify $\sqrt{6-\sqrt{20}}$ My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$
Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$
But I still end up with an ugly radical expression.

$$a=\sqrt{6-\sqrt{20}}$$
$$\implies a^2=6-\sqrt{20}$$
$$\implies(a^2-6)^2=20$$
$$\implies a^4-12a^2+16=0$$
$$\implies (a^2-2a-4)(a^2+2a-4)=0$$
so it will be one of the four possibilities $\pm\sqrt{5}\pm1$, and since $\sqrt{6-\sqrt{25}} \lt \sqrt{6-\sqrt{20}} \lt \sqrt{6-\sqrt{4}}$, you want the one which is in $[1,2]$ 
