## Geometry: Common Core (15th Edition)

$2 \sqrt {15}$
First, we rewrite the problem in the form of a fraction: $\sqrt {300} \div \sqrt {5} = \frac{\sqrt {300}}{\sqrt 5}$ Rewrite both numbers under one radical: $\sqrt {\frac{300}{5}}$ Evaluate what is under the radical sign: $\sqrt {60}$ Rewrite $60$ as the product of a perfect square and another number: $\sqrt {4 • 15}$ Take the square root of $4$ to solve: $2 \sqrt {15}$