[Python] Given A and B, find the number of all integer X values such that X*(X+1) falls...

Given A and B, find the number of all integer X values such that X*(X+1) falls between [A, B], both inclusive.

My approach:

X^2 + X >= A --> X^2 + X - A >= 0

X^2 + X <= B --> X^2 + X - B <= 0

1. 
   Find the roots of both equations using find_quadratic_roots() below.
   
   
2. 
   Use the higher and lower roots of both equations to derive all positive & negative X integers (this is where I am stuck)
   

from math import sqrt, floor, ceil

def find_quadratic_roots(a: int, b: int, c: int) -> tuple:
""" Returns roots of ax² + bx + c = 0
For our case, a and b will always be 1 """

discriminant = b*b - 4*a*c
if discriminant < 0:
return None

root1 = (-b + sqrt(discriminant)) / (2*a)
root2 = (-b - sqrt(discriminant)) / (2*a)

return (min(root1, root2), max(root1, root2))


def solution(A: int, B: int) -> int:
if A > B: return 0

# For lower bound: x² + x - A ≥ 0
lower_roots = find_quadratic_roots(1, 1, -A)

# For upper bound: x² + x - B ≤ 0
upper_roots = find_quadratic_roots(1, 1, -B)


assert solution(-1, 0) == 2 # -1, 0
assert solution(0, 0) == 2 # -1, 0
assert solution(0, 1) == 2 # -1, 0
assert solution(0, 2) == 4 # -2, -1, 0, 1
assert solution(-5, 5) == 4 # -2, -1, 0, 1
assert solution(0, 6) == 6 # -3, -2, -1, 0, 1, 2
assert solution(6, 6) == 2 # -3, 2
assert solution(3, 6) == 2 # -3, 2

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