Answer
$\dfrac{3}{\sqrt{7}-4}=-\dfrac{\sqrt{7}+4}{3}$
Work Step by Step
$\dfrac{3}{\sqrt{7}-4}$
Multiply the numerator and the denominator of this expression by the conjugate of the denominator and simplify:
$\dfrac{3}{\sqrt{7}-4}=\dfrac{3}{-4+\sqrt{7}}\cdot\dfrac{-4-\sqrt{7}}{-4-\sqrt{7}}=\dfrac{3(-4-\sqrt{7})}{(-4)^{2}-(\sqrt{7})^{2}}=...$
$...=\dfrac{3(-4-\sqrt{7})}{16-7}=\dfrac{3(-4-\sqrt{7})}{9}=\dfrac{-4-\sqrt{7}}{3}=-\dfrac{\sqrt{7}+4}{3}$
