## Algebra: A Combined Approach (4th Edition)

$x=-\dfrac{1}{6}\pm\dfrac{\sqrt{19}}{6}$
$x(6x+2)=3$ Evaluate the product on the left side of the equation: $6x^{2}+2x=3$ Take the $3$ to the left side: $6x^{2}+2x-3=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here $a=6$, $b=2$ and $c=-3$. Substitute: $x=\dfrac{-2\pm\sqrt{2^{2}-4(6)(-3)}}{2(6)}=\dfrac{-2\pm\sqrt{4+72}}{12}=...$ $...=\dfrac{-2\pm\sqrt{76}}{12}=\dfrac{-2\pm2\sqrt{19}}{12}=-\dfrac{1}{6}\pm\dfrac{\sqrt{19}}{6}$