Answer
$y=\dfrac{3}{2}\pm\dfrac{\sqrt{11}}{2}$
Work Step by Step
Using the completing the square method, then,
\begin{array}{l}
4y^2-2=12y\\
4y^2-12y-2=0\\
y^2-3y-\dfrac{1}{2}=0\\
y^2-3y=\dfrac{1}{2}\\
y^2-3y+\left( \dfrac{-3}{2} \right)^2=\dfrac{1}{2}+\left( \dfrac{-3}{2}\right)^2\\
y^2-3y+\dfrac{9}{4}=\dfrac{1}{2}+\dfrac{9}{4}\\
\left( y-\dfrac{3}{2} \right)^2=\dfrac{11}{4}\\
y-\dfrac{3}{2}=\pm\sqrt{\dfrac{11}{4}}\\
y=\dfrac{3}{2}\pm\dfrac{\sqrt{11}}{2}
.\end{array}
