#### Answer

$\dfrac{45-15\sqrt{5}}{4}$

#### Work Step by Step

Multiplying by the conjugate of the denominator, then
\begin{array}{l}\require{cancel}
\dfrac{15}{3+\sqrt{5}}
\\\\=
\dfrac{15}{3+\sqrt{5}}\cdot\dfrac{3-\sqrt{5}}{3-\sqrt{5}}
\\\\=
\dfrac{15(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}
.\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{15(3)+15(-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}
\\\\=
\dfrac{45-15\sqrt{5}}{(3+\sqrt{5})(3-\sqrt{5})}
.\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
\begin{array}{l}\require{cancel}
\dfrac{45-15\sqrt{5}}{(3+\sqrt{5})(3-\sqrt{5})}
\\\\=
\dfrac{45-15\sqrt{5}}{(3)^2-(\sqrt{5})^2}
\\\\=
\dfrac{45-15\sqrt{5}}{9-5}
\\\\=
\dfrac{45-15\sqrt{5}}{4}
.\end{array}