Answer
$u=-1+\sqrt{5} $ or $u=-1-\sqrt{5}$
Work Step by Step
$ u^{2}+2u-4=0\qquad$... solve with the Quadractic formula. $a=1,\ b=2,\ c=-4$
$ u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $2,\ a$ for $1$ and $c$ for $-4$.
$ u=\displaystyle \frac{-2\pm\sqrt{(2)^{2}-4\cdot(-4)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$u=\displaystyle \frac{-2\pm\sqrt{4+16}}{2}$
$u=\displaystyle \frac{-2\pm\sqrt{4\cdot 5}}{2}$
$u=\displaystyle \frac{-2\pm 2\sqrt{5}}{2}$
$ u=-1\pm\sqrt{5}\qquad$... the symbol $\pm$ indicates two solutions.
$u=-1+\sqrt{5} $ or $u=-1-\sqrt{5}$
