Solve by completing the square: \( 2 y^{2}+10 y=-11 \)

To solve the equation \( 2y^{2} + 10y = -11 \) by completing the square, we first move all terms to one side: \[ 2y^{2} + 10y + 11 = 0 \] Next, we divide every term by 2 to make the coefficient of \( y^{2} \) equal to 1: \[ y^{2} + 5y + \frac{11}{2} = 0 \] Now, we focus on the \( y^{2} + 5y \) part. To complete the square, take half of the coefficient of \( y \) (which is 5), square it, and add it to both sides. Half of 5 is \( \frac{5}{2} \), and squaring it gives \( \left(\frac{5}{2}\right)^{2} = \frac{25}{4} \): 1. Add \( \frac{25}{4} \) to both sides: \[ y^{2} + 5y + \frac{25}{4} = \frac{11}{2} + \frac{25}{4} \] 2. Convert \( \frac{11}{2} \) to a fraction with a denominator of 4: \[ \frac{11}{2} = \frac{22}{4} \] So now we have: \[ y^{2} + 5y + \frac{25}{4} = \frac{22}{4} + \frac{25}{4} \] This simplifies to: \[ y^{2} + 5y + \frac{25}{4} = \frac{47}{4} \] 3. The left side is a perfect square: \[ \left(y + \frac{5}{2}\right)^{2} = \frac{47}{4} \] 4. Taking the square root of both sides: \[ y + \frac{5}{2} = \pm \sqrt{\frac{47}{4}} \] 5. Simplifying further: \[ y + \frac{5}{2} = \pm \frac{\sqrt{47}}{2} \] 6. Finally, solving for \( y \): \[ y = -\frac{5}{2} \pm \frac{\sqrt{47}}{2} \] Which gives us the two solutions: \[ y = \frac{-5 + \sqrt{47}}{2} \quad \text{and} \quad y = \frac{-5 - \sqrt{47}}{2} \] And there you have it, the complete solutions after completing the square!