$\dfrac{\sqrt{6}+3\sqrt{2}}{3+\sqrt{5}}=\dfrac{ \sqrt{6}+3\sqrt{2}}{3+\sqrt{5}}\cdot 1=$

$\dfrac{\sqrt{6}+3\sqrt{2}}{3+\sqrt{5}}\cdot \dfrac{3-\sqrt{5}}{3-\sqrt{5}}=$

$\dfrac{3\sqrt{6}+9\sqrt{2}-\sqrt{30}-3\sqrt{10}}{9-5}=$

$\dfrac{3\sqrt{6}+9\sqrt{2}-\sqrt{30}-3\sqrt{10}}{4}$

this form should be acceptable. You can tinker with it further to get

$\dfrac{3\sqrt{2}\sqrt{3}+9\sqrt{2}-\sqrt{2}\sqrt{3}\sqrt{5}-3\sqrt{2}\sqrt{5}}{4}=$

$\dfrac{\sqrt{2} \left(3\sqrt{3}+9-\sqrt{3} \sqrt{5}-3\sqrt{5}\right)}{4}=$

$\dfrac{\sqrt{2}\left(3+\sqrt{3}\right)\left(3-\sqrt{5}\right)}{4}$