## College Algebra (10th Edition)

The solution set is {$\frac{3-3i\sqrt2}{4}, \frac{3+3i\sqrt2}{4}$}
$4u^2-6u+9=0$ $a=4, b=-6, c=9$ $D=b^2-4ac=(-6)^2-4\cdot4\cdot9=36-144=-108<0$. The equation has two complex solutions. $u=\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{6\pm\sqrt {-108}}{2\cdot4}$ Use the quadratic formula. $u=\frac{6\pm\sqrt {-3^2{\cdot2^2\cdot2}}}{2\cdot4}$ $u=\frac{6\pm6i\sqrt2}{8}$ $u=\frac{3\pm3i\sqrt2}{4}$ The solution set is {$\frac{3-3i\sqrt2}{4}, \frac{3+3i\sqrt2}{4}$}