## Precalculus (6th Edition)

The solutions are $x=-\dfrac{1}{3}\pm\dfrac{\sqrt{14}}{3}i$
$-x(3x+2)=5$ Evaluate the product on the left side: $-3x^{2}-2x=5$ Take all terms to the right side and rearrange: $0=3x^{2}+2x+5$ $3x^{2}+2x+5=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=3$, $b=2$ and $c=5$. Substitute the known values into the formula and evaluate: $x=\dfrac{-2\pm\sqrt{2^{2}-4(3)(5)}}{2(3)}=\dfrac{-2\pm\sqrt{4-60}}{6}=\dfrac{-2\pm\sqrt{-56}}{6}=...$ $...=\dfrac{-2\pm2\sqrt{14}i}{6}=-\dfrac{2}{6}\pm\dfrac{2\sqrt{14}}{6}i=-\dfrac{1}{3}\pm\dfrac{\sqrt{14}}{3}i$ The solutions are $x=-\dfrac{1}{3}\pm\dfrac{\sqrt{14}}{3}i$