Answer
$\dfrac{7+\sqrt{3}}{23}$
Work Step by Step
Multiplying by the conjugate of the denominator, the rationalized-denominator from of the given expression, $
\dfrac{2}{7-\sqrt{3}}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{2}{7-\sqrt{3}}\cdot\dfrac{7+\sqrt{3}}{7+\sqrt{3}}
\\\\=
\dfrac{2(7+\sqrt{3})}{(7)^2-(\sqrt{3})^2}
\\\\=
\dfrac{2(7+\sqrt{3})}{49-3}
\\\\=
\dfrac{2(7+\sqrt{3})}{46}
\\\\=
\dfrac{\cancel{2}(7+\sqrt{3})}{\cancel{2}(23)}
\\\\=
\dfrac{7+\sqrt{3}}{23}
.\end{array}
