Algebra 1

$\frac{3\sqrt{7}+3\sqrt{3}}{4}$
Multiply using the conjugate. Remember that $(a-b)(a+b)=a^2-b^2$ $\frac{3}{\sqrt{7}-\sqrt{3}}\cdot \frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}}=\frac{3\sqrt{7}+3\sqrt{3}}{7-3}=\frac{3\sqrt{7}+3\sqrt{3}}{4}$