## Precalculus (6th Edition)

The solutions are $x=\dfrac{3}{2}\pm\dfrac{\sqrt{2}}{2}i$
$-4x^{2}=-12x+11$ Take all terms to the left side of the equation: $-4x^{2}+12x-11=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=-4$, $b=12$ and $c=-11$ Substitute the known values into the formula and evaluate: $x=\dfrac{-12\pm\sqrt{12^{2}-4(-4)(-11)}}{2(-4)}=\dfrac{-12\pm\sqrt{144-176}}{-8}=...$ $...=\dfrac{-12\pm\sqrt{-32}}{-8}=\dfrac{-12\pm4\sqrt{2}i}{-8}=\dfrac{-12}{-8}\pm\dfrac{4\sqrt{2}}{8}i=...$ $...=\dfrac{3}{2}\pm\dfrac{\sqrt{2}}{2}i$ The solutions are $x=\dfrac{3}{2}\pm\dfrac{\sqrt{2}}{2}i$