Answer
$y=\dfrac{-3\pm\sqrt{21}}{3}$
Work Step by Step
Using the completing the square method, the solutions of the given quadratic equation, $
3y^2+6y-4=0
,$ is
\begin{array}{l}\require{cancel}
\dfrac{1}{3}\cdot(3y^2+6y-4)=(0)\cdot\dfrac{1}{3}
\\\\
y^2+2y-\dfrac{4}{3}=0
\\\\
y^2+2y=\dfrac{4}{3}
\\\\
y^2+2y+\left(\dfrac{2}{2} \right)^2=\dfrac{4}{3}+\left(\dfrac{2}{2} \right)^2
\\\\
y^2+2y+1=\dfrac{4}{3}+1
\\\\
(y+1)^2=\dfrac{7}{3}
\\\\
y+1=\pm\sqrt{\dfrac{7}{3}}
\\\\
y+1=\pm\dfrac{\sqrt{21}}{3}
\\\\
y=-1\pm\dfrac{\sqrt{21}}{3}
\\\\
y=-\dfrac{3}{3}\pm\dfrac{\sqrt{21}}{3}
\\\\
y=\dfrac{-3\pm\sqrt{21}}{3}
.\end{array}
