Answer
$y^2-5y+\frac{25}{4} = (y-\frac{5}{2})^2$
Work Step by Step
We add the square of half of the co-efficient of y
Co-efficient of y = -5
Half of -5 = $\frac{1}{2} \times -5 = \frac{-5}{2}$
Square of $\frac{-5}{2} = \frac{-5}{2}\times\frac{-5}{2} = \frac{25}{4}$
We add $\frac{25}{4}$ to make $y^2-5y$ a perfect square trinomial.
Therefore, $y^2-5y+\frac{25}{4}$ is a perfect square trinomial.
Factor form of $y^2-5y+\frac{25}{4}$
= $(y - \frac{5}{2})^2$
