Answer
$\left\{ -1-\sqrt{19},-1+\sqrt{19} \right\}$
Work Step by Step
The standard form of the given equation, $
x(6x+2)=3
,$ is
\begin{array}{l}\require{cancel}
6x^2+2x=3
\\\\
6x^2+2x-3=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
6x^2+2x-3=0
,$ are
\begin{array}{l}\require{cancel}
\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\pm\sqrt{4\cdot19}}{12}
\\\\=
\dfrac{-2\pm\sqrt{(2)^2\cdot19}}{12}
\\\\=
\dfrac{-2\pm2\sqrt{19}}{12}
\\\\=
\dfrac{2(-1\pm\sqrt{19})}{12}
\\\\=
\dfrac{\cancel{2}(-1\pm\sqrt{19})}{\cancel{2}\cdot6}
\\\\=
-1\pm\sqrt{19}
.\end{array}
Hence, the solutions are $
\left\{ -1-\sqrt{19},-1+\sqrt{19} \right\}
.$
