Answer
$u=1+\sqrt{3}$ or $u=1-\sqrt{3}$
Work Step by Step
$ u^{2}-2u-2=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-2,\ c=-2$
$ u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-2,\ a$ for $1$ and $c$ for $-2$.
$ u=\displaystyle \frac{-(-2)\pm\sqrt{(-2)^{2}-4\cdot(-2)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$u=\displaystyle \frac{2\pm\sqrt{4+8}}{2}$
$u=\displaystyle \frac{2\pm\sqrt{12}}{2}$
$u=\displaystyle \frac{2\pm\sqrt{4\cdot 3}}{2}$
$u=\displaystyle \frac{2\pm 2\sqrt{3}}{2}$
$ u=1\pm\sqrt{3}\qquad$... the symbol $\pm$ indicates two solutions.
$u=1+\sqrt{3}$ or $u=1-\sqrt{3}$
