The quadratic equation is given as
[tex](y+\frac{1}{3})^2=81[/tex]
Using the square roots property, we need to find the square root of both sides.
Therefore, we get
[tex]y+\frac{1}{3}=\pm\sqrt[]{81}[/tex]
Hence, we can solve as
[tex]\begin{gathered} y+\frac{1}{3}=\pm9 \\ y=\pm9-\frac{1}{3} \end{gathered}[/tex]
The roots can be gotten as
[tex]\begin{gathered} y=9-\frac{1}{3} \\ y=8\frac{2}{3} \end{gathered}[/tex]
or
[tex]\begin{gathered} y=-9-\frac{1}{3} \\ y=-9\frac{1}{3} \end{gathered}[/tex]
The answer is
[tex]y=8\frac{2}{3},-9\frac{1}{3}[/tex]
