Solve the quadratic equation by completing the square. \( 3 y^{2}-4 y-11=0 \)

To solve the equation \( 3 y^{2} - 4 y - 11 = 0 \) by completing the square, first divide every term by 3 to simplify: \[ y^{2} - \frac{4}{3} y - \frac{11}{3} = 0 \] Next, move the constant term to the other side: \[ y^{2} - \frac{4}{3} y = \frac{11}{3} \] Now, we need to complete the square. Take half of the coefficient of \( y \), square it, and add it to both sides. Half of \(-\frac{4}{3}\) is \(-\frac{2}{3}\), and squaring it gives \(\frac{4}{9}\): \[ y^{2} - \frac{4}{3} y + \frac{4}{9} = \frac{11}{3} + \frac{4}{9} \] On the left side, we can now express it as a square: \[ \left( y - \frac{2}{3} \right)^{2} \] Now, we need to simplify the right side. Convert \(\frac{11}{3}\) to a fraction with a denominator of 9: \[ \frac{11}{3} = \frac{33}{9} \] Thus, you add: \[ \frac{33}{9} + \frac{4}{9} = \frac{37}{9} \] So we have: \[ \left( y - \frac{2}{3} \right)^{2} = \frac{37}{9} \] Taking the square root of both sides gives: \[ y - \frac{2}{3} = \pm \frac{\sqrt{37}}{3} \] Finally, solve for \( y \): \[ y = \frac{2}{3} \pm \frac{\sqrt{37}}{3} \] So, the solutions are: \[ y = \frac{2 + \sqrt{37}}{3} \quad \text{and} \quad y = \frac{2 - \sqrt{37}}{3} \]