College Algebra (6th Edition)

The solutions are $x=\dfrac{1}{5}\pm\dfrac{\sqrt{14}}{5}i$
$5x^{2}=2x-3$ Take all terms to the right side of the equation: $5x^{2}-2x+3=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ In this case, $a=5$, $b=-2$ and $c=3$ Substitute the known values into the formula and evaluate: $x=\dfrac{-(-2)\pm\sqrt{(-2)^{2}-4(5)(3)}}{2(5)}=\dfrac{2\pm\sqrt{4-60}}{10}=...$ $...=\dfrac{2\pm\sqrt{-56}}{10}=\dfrac{2\pm2\sqrt{14}i}{10}=\dfrac{1}{5}\pm\dfrac{\sqrt{14}}{5}i$