# How do you solve p^2 - 4p + 4 = 0  using the quadratic formula?

The unique (double-root) in this case is $p = 2$.
For an equation of the form $a {x}^{2} + b x + c$, the quadratic formula says the solutions are $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$. For this problem, $a = 1$, $b = - 4$, and $c = 4$. Thus, there's just one answer:
$p = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{4 \pm 0}{2} = 2$.