Denest $\sqrt{\left(2+\sqrt{2}\right) \left(10-2 \sqrt{5}\right)}$ Is it possible to denest following radical to sum of terms with smaller root count inside?

$\sqrt{20+10 \sqrt{2}-4 \sqrt{5}-2 \sqrt{10}}=\sqrt{\left(2+\sqrt{2}\right) \left(10-2 \sqrt{5}\right)}$

I've found that it cannot be denested into $a+b\sqrt{2}+c\sqrt{5}+d\sqrt{10}$ by squaring and equating coefficients. Resulting system didn't have rational solutions.

Maybe it can be denested to something as sum of $\sqrt{a+b\sqrt{c}}$ or $\sqrt{a+b\sqrt{c}+d\sqrt{e}}$ or even $\sqrt[4]{a+b\sqrt{c}}$?

For example, following can be denested:
$$\sqrt{8+2 \sqrt{2}-2 \sqrt{6}}=\sqrt{6+3\sqrt{2}}-\sqrt{2-\sqrt{2}}$$

$$\sqrt{110-60 \sqrt{3}+46 \sqrt{5}-28 \sqrt{15}}=2\sqrt{15+6\sqrt{5}}-\sqrt{50+22\sqrt{5}}$$

$$\sqrt{48+12 \sqrt{2}+16 \sqrt{5}-12 \sqrt{6}+4\sqrt{10}-4\sqrt{30}}=$$
$$=\sqrt{6+3\sqrt{2}}-\sqrt{2-\sqrt{2}}+\sqrt{30+15\sqrt{2}}-\sqrt{10-5\sqrt{2}}$$
Well if you're still looking for something:

$$\sqrt{2+\sqrt{2}}=\sqrt{\frac{2+\sqrt{2}}{2}}+\sqrt{\frac{2-\sqrt{2}}{2}}$$
and

$$\sqrt{10-2\sqrt{5}}=2 \cdot \sqrt{\frac{5-\sqrt{5}}{2}}$$

so
$$\sqrt{\left(2+\sqrt{2}\right) \cdot \left(10-2\sqrt{5}\right)}=2\cdot \sqrt{\frac{5-\sqrt{5}}{2}} \cdot \left(\sqrt{\frac{2+\sqrt{2}}{2}}+\sqrt{\frac{2-\sqrt{2}}{2}}\right)$$
Try factoring $2\sqrt{10}$ as $2\sqrt{5}\sqrt{2}$, then the expression under the radical can be factored.