We are asked to solve the quadratic equation $2y^2 + 10y = -9$ by completing the square.

1. Problem Description

We are asked to solve the quadratic equation $2y^2 + 10y = -9$ by completing the square.

2. Solution Steps

First, we want the coefficient of the $y^2$ term to be

1. We divide both sides of the equation by 2:

$y^2 + 5y = -\frac{9}{2}$

To complete the square, we need to add $(\frac{5}{2})^2 = \frac{25}{4}$ to both sides of the equation:

$y^2 + 5y + \frac{25}{4} = -\frac{9}{2} + \frac{25}{4}$

The left side is now a perfect square: $(y + \frac{5}{2})^2$. The right side simplifies to:

$-\frac{9}{2} + \frac{25}{4} = -\frac{18}{4} + \frac{25}{4} = \frac{7}{4}$

So we have:

$(y + \frac{5}{2})^2 = \frac{7}{4}$

Now, we take the square root of both sides:

$y + \frac{5}{2} = \pm \sqrt{\frac{7}{4}} = \pm \frac{\sqrt{7}}{2}$

Finally, we solve for $y$:

$y = -\frac{5}{2} \pm \frac{\sqrt{7}}{2} = \frac{-5 \pm \sqrt{7}}{2}$

3. Final Answer

$\frac{-5 + \sqrt{7}}{2}, \frac{-5 - \sqrt{7}}{2}$