Assume we have the equation:
$$c = (a-n)(b-n)$$
where a, b and c are constants.
Solving for n, we find:
$$n = \frac{a+b \pm \sqrt{a^2 -2ab + b^2 + 4c} }{2}$$
Let z be the the minimum absolute value of the roots. Can we prove 0 < z < a and 0 < z < b for all n?
For example, let a=150, b=100 and c=10,000, then:
$$10,000 = (150-n)(100-n)$$
The roots are:
$$n=228.07...$$
and
$$n=21.92...$$
Then:
$$z = min(abs(228.07), abs(21.92)) = 21.92$$
By brute force I see 0 < z < a and 0 < z < b for all n, but I can't prove it analytically
Can we prove that $0 < z < a$ and $0 < z < b$ for all $n$?
I guess you meant to ask: Can we prove that $0 < z < a$ and $0 < z < b$ for all $c$?
It is not possible for all $c$.
If $c = 0$, then we have $$ (n-a)(n-b) = 0 $$ and the two roots are $n = a$ and $n = b$.