Answer
$\dfrac{6+2\sqrt{2}+3\sqrt{3}+\sqrt{6}}{7}$
Work Step by Step
Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2,$ the given expression, $
\dfrac{2+\sqrt{3}}{3-\sqrt{2}}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{2+\sqrt{3}}{3-\sqrt{2}}\cdot\dfrac{3+\sqrt{2}}{3+\sqrt{2}}
\\\\=
\dfrac{2(3)+2(\sqrt{2})+\sqrt{3}(3)+\sqrt{3}(\sqrt{2})}{(3)^2-(\sqrt{2})^2}
\\\\=
\dfrac{6+2\sqrt{2}+3\sqrt{3}+\sqrt{6}}{9-2}
\\\\=
\dfrac{6+2\sqrt{2}+3\sqrt{3}+\sqrt{6}}{7}
.\end{array}
