Simplify the fraction with radicals I want to simplify this fraction
$$ \frac{\sqrt{6} + \sqrt{10} + \sqrt{15} + 2}{\sqrt{6} - \sqrt{10} + \sqrt{15} - 2} $$
I've tried to group up the denominator members like $ (\sqrt{6} + \sqrt{15}) - (\sqrt{10} + 2) $ and then amplify with $ (\sqrt{6} + \sqrt{15}) + (\sqrt{10} + 2) $ 

For this system we implemented finding a split form for the denomiator:
$$a + \sqrt{p}\,b$$
Such that $\sqrt{p}$ is a new radical. For a quotient we then have:
$$\frac{c}{a + \sqrt{p}\,b} = \frac{c\,(a - \sqrt{p}\,b)}{a^2 - p\,b^2}$$
Lets give it a try:
$$\frac{2+\sqrt{6}+\sqrt{10}+\sqrt{15}}{-2+\sqrt{6}-\sqrt{10}+\sqrt{15}} = $$
$$\frac{2+\sqrt{15}+\sqrt{2}(\sqrt{3}+\sqrt{5})}{- 2+\sqrt{15}+\sqrt{2}(\sqrt{3}-\sqrt{5})} =$$
$$\frac{(2+\sqrt{15}+\sqrt{2}(\sqrt{3}+\sqrt{5}))\,(- 2+\sqrt{15}-\sqrt{2}(\sqrt{3}-\sqrt{5}))}{(- 2+\sqrt{15})^2-2(\sqrt{3}-\sqrt{5})^2} = $$
$$\frac{15+6\sqrt{6}}{3} = 5+2\sqrt{6}$$
