Answer
$$r=4,\:r=2$$
Work Step by Step
Completing the square, we find:
$$ r^2-6r+8+\left(-3\right)^2-\left(-3\right)^2 \\ \left(r-3\right)^2+8-\left(-3\right)^2 \\ \left(r-3\right)^2-1$$
Setting this equal to zero:
$$ \left(r-3\right)^2=1 \\ r=4,\:r=2$$