Answer
$y=\pm\sqrt{3}$
Work Step by Step
Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Principle), the solutions to the given equation, $
4y^2=12
,$ are
\begin{array}{l}\require{cancel}
y^2=\dfrac{12}{4}
\\\\
y^2=3
\\\\
y=\pm\sqrt{3}
.\end{array}