## Functions Modeling Change: A Preparation for Calculus, 5th Edition

$$q=\frac{\sqrt{41}+1}{5},\:q=\frac{-\sqrt{41}+1}{5}$$
Completing the square, we find: $$5\left(q^2-\frac{2q}{5}-\frac{8}{5}\right) \\ 5\left(q-\frac{1}{5}\right)^2-\frac{41}{5}$$ Setting this equal to zero: $$5\left(q-\frac{1}{5}\right)^2=\frac{41}{5} \\ q=\frac{\sqrt{41}+1}{5},\:q=\frac{-\sqrt{41}+1}{5}$$