Answer: \(y = -2 \text{ or } y = -\frac{5}{2}\)
Divide by 2. Half of \(\frac{9}{2}\) is \(\frac{9}{4}\).
\[y^2 + \frac{9}{2}y + 5 = 0 \quad (\div 2)\]\[y^2 + \frac{9}{2}y + \frac{81}{16} = \frac{81}{16} - 5\]\[\left(y + \frac{9}{4}\right)^2 = \frac{1}{16}\]\[y + \frac{9}{4} = \pm\frac{1}{4}\]\(y = -\frac{9}{4} + \frac{1}{4} = -2\) or \(y = -\frac{9}{4} - \frac{1}{4} = -\frac{5}{2}\).