Answer: \(y = \frac{2 + \sqrt{2}}{3} \text{ or } y = \frac{2 - \sqrt{2}}{3}\)
Make the coefficient of \(y^2\) equal to 1 first. Half of \(-\frac{4}{3}\) is \(-\frac{2}{3}\).
\[y^2 - \frac{4}{3}y + \frac{2}{9} = 0 \quad (\div 9)\]\[y^2 - \frac{4}{3}y + \frac{4}{9} = \frac{4}{9} - \frac{2}{9}\]\[\left(y - \frac{2}{3}\right)^2 = \frac{2}{9}\]\[y - \frac{2}{3} = \pm\frac{\sqrt{2}}{3}\]\(y = \frac{2 + \sqrt{2}}{3}\) or \(y = \frac{2 - \sqrt{2}}{3}\).