real analysis – Necessity of selecting value for $delta$ in $epsilon$-$delta$ proofs

I’ve been reviewing the epsilon-delta definition of a limit before beginning Analysis 1 and am slightly confused about the problems where $delta$ is selected from a minimum of two values.

For example, a problem in my textbook asks to prove $$lim_{x to 3} x^2 = 9$$

using epsilon-delta. The process is explained as follows:

$$|x^2-9|<epsilon$$

$$|x+3||x-3|<epsilon$$

$$text{Let } delta = 1, text{so } |x-3|<1$$

$$-1<x-3<1$$

$$5<x+3<7$$

$$therefore |x-3|<frac{epsilon}{7}$$

$$delta = min { 1, frac{epsilon}{7} }$$

I’m confused as to why we don’t leave delta as delta, and simply solve for delta in terms of epsilon (i.e. $delta = sqrt{epsilon+9}-3$). Are we able to simply pick a bound for $delta$ because epsilon-delta proofs only require that there exists a positive delta that will work, and that there an infinite number of positive deltas that exist in $left(0,1right)$, for example? Otherwise, the proof seems incomplete because it doesn’t take into account any $delta>1$.