>>11360399the parabola is symmetric. and zeros are products of values of the form (x + a).
we can simply set our value to (x + d + e/2)(x + d - e/2) where 'd' offsets us to the middle of the parabola and '±e/2' should take us to either zeroes. consider some arbitrary parabola. i will assume the highest degree term has a constant 1 (this is fine as polynomials share zeroes up to scalar multiplications):
x^2 + Bx + C = 0
and putting them in a single equation we get:
(x + d + e/2)(x + d - e/2) = x^2 + Bx + C
now, (x + d + e/2)(x + d - e/2) = x^2 + 2dx + d^2 - (e/2)^2
{x^2, x, 1} are linear independent functions so we can then extract a system of equations. mainly:
d = B/2
C = d^2 - (e/2)^2 = (B/2)^2 - (e/2)^2 => (e/2)^2 = (B/2)^2 - C => e/2 = sqrt((B/2)^2 - C) = sqrt(B^2 - 4C)/2
we then get that x = -B/2 ± sqrt(B^2 - 4C)/2
note how there is now A as ive set A = 1 though doing the same thing with this condition shouldn't be too difficult!
