The given equation is:
[tex]3y^2=432[/tex]Divide both sides of the equation by 3:
[tex]\begin{gathered} \frac{3y^2}{3}=\frac{432}{3} \\ y^2=144 \end{gathered}[/tex]Therefore,
[tex]\begin{gathered} y=\pm\sqrt{144} \\ y=\pm12 \end{gathered}[/tex]Hence
y = -12 or y = 12
