To find the perfect square trinomial, we proceed as follows;
[tex]\begin{gathered} y^2\text{ + }\frac{1}{2}y\text{ = -}\frac{1}{16} \\ \\ y^2\text{ + }\frac{1}{2}y\text{ + }\frac{1}{16}\text{ = 0} \\ \\ y^2\text{ + }\frac{1}{2}y\text{ + (}\frac{1}{4})^2\text{ = 0} \\ \\ =(\text{ y + }\frac{1}{4})^2\text{ = 0} \\ \\ (y\text{ + }\frac{1}{4})(y\text{ + }\frac{1}{4})\text{ = 0} \\ \\ y\text{ = -}\frac{1}{4}\text{ twice} \end{gathered}[/tex]