Answer
$-4\sqrt{10}+4\sqrt{15}$
Work Step by Step
Multiplying by the conjugate of the denominator, the given expression is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{4\sqrt{5}}{\sqrt{2}+\sqrt{3}}
\\\\=
\dfrac{4\sqrt{5}}{\sqrt{2}+\sqrt{3}}\cdot\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}
\\\\=
\dfrac{4\sqrt{5}(\sqrt{2}-\sqrt{3})}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}
.\end{array}
Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{4\sqrt{5}(\sqrt{2})+4\sqrt{5}(-\sqrt{3})}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}
\\\\=
\dfrac{4\sqrt{5(2)}-4\sqrt{5(3)}}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}
\\\\=
\dfrac{4\sqrt{10}-4\sqrt{15}}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}
.\end{array}
Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{4\sqrt{10}-4\sqrt{15}}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}
\\\\=
\dfrac{4\sqrt{10}-4\sqrt{15}}{(\sqrt{2})^2-(\sqrt{3})^2}
\\\\=
\dfrac{4\sqrt{10}-4\sqrt{15}}{2-3}
\\\\=
\dfrac{4\sqrt{10}-4\sqrt{15}}{-1}
\\\\=
-4\sqrt{10}+4\sqrt{15}
.\end{array}
