Answer
$\dfrac{\sqrt{3}}{3} \text{ }$
Work Step by Step
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the given quadratic equation, $
3x^2-\sqrt{12}x+1=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-\sqrt{12})\pm\sqrt{(-\sqrt{12})^2-4(3)(1)}}{2(3)}
\\\\=
\dfrac{\sqrt{12}\pm\sqrt{12-12}}{6}
\\\\=
\dfrac{\sqrt{12}\pm\sqrt{0}}{6}
\\\\=
\dfrac{\sqrt{12}\pm0}{6}
\\\\=
\dfrac{\sqrt{12}}{6}
\\\\=
\dfrac{\sqrt{4\cdot3}}{6}
\\\\=
\dfrac{\sqrt{(2)^2\cdot3}}{6}
\\\\=
\dfrac{2\sqrt{3}}{6}
\\\\=
\dfrac{\cancel{2}\sqrt{3}}{\cancel{2}\cdot3}
\\\\=
\dfrac{\sqrt{3}}{3}
.\end{array}
Hence, the solutions are $
\dfrac{\sqrt{3}}{3} \text{ }
.$
