Answer
$\dfrac{12-2\sqrt{3}+6\sqrt{5}-\sqrt{15}}{33}$
Work Step by Step
Multiplying by the conjugate of the denominator, the rationalized-denominator form of the given expression, $
\dfrac{2+\sqrt{5}}{6+\sqrt{3}}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{2+\sqrt{5}}{6+\sqrt{3}}\cdot\dfrac{6-\sqrt{3}}{6-\sqrt{3}}
\\\\=
\dfrac{2(6)+2(-\sqrt{3})+\sqrt{5}(6)+\sqrt{5}(-\sqrt{3})}{6^2-(\sqrt{3})^2}
\\\\=
\dfrac{12-2\sqrt{3}+6\sqrt{5}-\sqrt{5(3)}}{36-3}
\\\\=
\dfrac{12-2\sqrt{3}+6\sqrt{5}-\sqrt{15}}{33}
.\end{array}
