Answer
the solution set is\[\left\{ \frac{-3\pm \sqrt{3}}{6} \right\}\].
Work Step by Step
The given equation can be written as\[6{{x}^{2}}+6x+1=0\].
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\].
Here,\[a=6,\,\,b=6,\text{ and }c=1\].
Now put these values in the quadratic formula: \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
That is,
\[\begin{align}
& x=\frac{-6\pm \sqrt{{{\left( 6 \right)}^{2}}-4\times 6\times 1}}{2\times 6} \\
& =\frac{-6\pm \sqrt{36-24}}{12} \\
& =\frac{-6\pm \sqrt{12}}{12} \\
& =\frac{-6\pm 2\sqrt{3}}{12}
\end{align}\]
Further simplify it to get,
\[\begin{align}
& x=\frac{-6\pm 2\sqrt{3}}{12} \\
& =2\left( \frac{-3\pm \sqrt{3}}{12} \right) \\
& =\frac{-3\pm \sqrt{3}}{6}
\end{align}\]
Hence, the solution set is\[\left\{ \frac{-3\pm \sqrt{3}}{6} \right\}\].
