## Algebra 2 (1st Edition)

The solutions are $\displaystyle \frac{-2+\sqrt{62}i}{6}$ and $\displaystyle \frac{-2-\sqrt{62}i}{6}$.
$6u^{2}+4u+11=0\qquad$ ...use the Quadratic formula: $x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $a=6,b=4,c=11$ $u=\displaystyle \frac{-4\pm\sqrt{16-264}}{12}\qquad$...simplify $u=\displaystyle \frac{-4\pm\sqrt{-248}}{12}\qquad$...rewrite using the imaginary unit $i$ $u=\displaystyle \frac{-4\pm 2\sqrt{62}i}{12}\qquad$...divide the expression with $2$. $u=\displaystyle \frac{-2\pm\sqrt{62}i}{6}$