Answer
$y=-\frac{1}{4}$.
Work Step by Step
The given polynomial is
$\Rightarrow y^2+\frac{1}{2}y=-\frac{1}{16}$
Add $\frac{1}{16}$ to each side.
$\Rightarrow y^2+\frac{1}{2}y+\frac{1}{16}=-\frac{1}{16}+\frac{1}{16}$
Simplify.
$\Rightarrow y^2+\frac{1}{2}y+\frac{1}{16}=0$
Write the the polynomial as $a^2+2ab+b^2$.
$\Rightarrow y^2+2(y)(\frac{1}{4})+(\frac{1}{4})^2=0$.
Use perfect square trinomial pattern
$a^2+2ab+b^2=(a+b)^2$.
We have $a=y$ and $b=\frac{1}{4}$.
$\Rightarrow (y+\frac{1}{4})^2=0$.
Use zero product property.
$\Rightarrow y+\frac{1}{4}=0$.
Solve for $y$.
$\Rightarrow y=-\frac{1}{4}$.
Hence, the solution is $y=-\frac{1}{4}$.
