## Precalculus: Mathematics for Calculus, 7th Edition

$x=\dfrac{3}{4}\pm\dfrac{\sqrt{7}}{4}i$
$2x^{2}-3x+2=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=2$, $b=-3$ and $c=2$. Substitute the known values into the formula and simplify: $x=\dfrac{-(-3)\pm\sqrt{(-3)^{2}-4(2)(2)}}{2(2)}=\dfrac{3\pm\sqrt{9-16}}{4}=...$ $...=\dfrac{3\pm\sqrt{-7}}{4}=\dfrac{3\pm\sqrt{7}i}{4}=\dfrac{3}{4}\pm\dfrac{\sqrt{7}}{4}i$