Answer
$\left\{ \dfrac{-3- i\sqrt{6}}{3},\dfrac{-3+i\sqrt{6}}{3} \right\}$
Work Step by Step
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the given quadratic equation, $
3y^2+6y+5
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(6)\pm\sqrt{(6)^2-4(3)(5)}}{2(3)}
\\\\=
\dfrac{-6\pm\sqrt{36-60}}{6}
\\\\=
\dfrac{-6\pm\sqrt{-24}}{6}
\\\\=
\dfrac{-6\pm\sqrt{-1}\sqrt{24}}{6}
\\\\=
\dfrac{-6\pm i\sqrt{4\cdot6}}{6}
\\\\=
\dfrac{-6\pm i\sqrt{(2)^2\cdot6}}{6}
\\\\=
\dfrac{-6\pm 2i\sqrt{6}}{6}
\\\\=
\dfrac{2(-3\pm i\sqrt{6})}{6}
\\\\=
\dfrac{\cancel{2}(-3\pm i\sqrt{6})}{\cancel{2}\cdot3}
\\\\=
\dfrac{-3\pm i\sqrt{6}}{3}
.\end{array}
Hence, the solutions are $
\left\{ \dfrac{-3- i\sqrt{6}}{3},\dfrac{-3+i\sqrt{6}}{3} \right\}
.$
