Answer
$x=-\dfrac{1}{6}\pm\dfrac{\sqrt{19}}{6}$
Work Step by Step
$x(6x+2)=3$
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:
$x=\dfrac{-2\pm\sqrt{2^{2}-4(6)(-3)}}{2(6)}=\dfrac{-2\pm\sqrt{4+72}}{12}=...$
$...=\dfrac{-2\pm\sqrt{76}}{12}=\dfrac{-2\pm2\sqrt{19}}{12}=-\dfrac{1}{6}\pm\dfrac{\sqrt{19}}{6}$
