## Algebra 1

$\frac{1+\sqrt{3}}{4}$
We must multiply by the conjugate (change the sign between the two terms ) of the denominator. $\frac{-1}{2-2\sqrt{3}}\cdot \frac{2+2\sqrt{3}}{2+2\sqrt{3}}=\frac{-2-2\sqrt{3}}{4+4\sqrt{3}-4\sqrt{3}-(2\sqrt{3})^2}=\frac{-2-2\sqrt{3}}{4-4(3)}=\frac{-2-2\sqrt{3}}{-8}=\frac{1+\sqrt{3}}{4}$