Quadratic Formula — Finding Where a Parabola Crosses Zero

Plain English first

A quadratic equation is any equation with an x² in it: ax² + bx + c = 0. Its graph is a parabola — a U-shape (or upside-down U). The solutions are wherever that parabola crosses the x-axis (where y = 0).

Sometimes a parabola crosses the x-axis twice (two solutions). Sometimes it just touches it once (one solution). Sometimes it doesn't reach the x-axis at all (no real solutions). The quadratic formula tells you which case you're in and what the answers are — without graphing anything.

The formula looks intimidating but it's just completing the square: rearranging the equation until one side is a perfect square, then taking the square root of both sides.

Standard math notation

For ax² + bx + c = 0:

       −b ± √(b² − 4ac)
  x = ——————————————————
              2a

The discriminant b² − 4ac tells you how many real solutions exist:
  b² − 4ac > 0 : two real solutions (parabola crosses x-axis twice)
  b² − 4ac = 0 : one real solution  (parabola just touches x-axis)
  b² − 4ac < 0 : no real solutions  (parabola doesn't reach x-axis)

The ± means you compute two answers:
  x₁ = (−b + √(b²−4ac)) / (2a)
  x₂ = (−b − √(b²−4ac)) / (2a)

Verbose Python with descriptive names

def solve_quadratic_equation(
    coefficient_of_x_squared,
    coefficient_of_x,
    constant_term
):
    """
    Find x values where ax² + bx + c = 0.
    These are the x-axis crossings of the parabola y = ax² + bx + c.

    The discriminant = b² - 4ac tells us how many real solutions exist.
    Formula: x = (-b ± √discriminant) / (2a)
    """
    a = coefficient_of_x_squared
    b = coefficient_of_x
    c = constant_term

    discriminant = b**2 - 4 * a * c

    if discriminant > 0:
        square_root_of_discriminant = discriminant ** 0.5
        first_solution  = (-b + square_root_of_discriminant) / (2 * a)
        second_solution = (-b - square_root_of_discriminant) / (2 * a)
        return [first_solution, second_solution]

    elif discriminant == 0:
        only_solution = -b / (2 * a)
        return [only_solution]

    else:  # discriminant < 0
        return []  # no real solutions — parabola never reaches x-axis

# x² - 5x + 6 = 0  →  factors as (x-2)(x-3)  →  x=2 or x=3
solutions = solve_quadratic_equation(
    coefficient_of_x_squared=1,
    coefficient_of_x=-5,
    constant_term=6
)
print(f"Solutions: {solutions}")  # [3.0, 2.0]

# x² + 1 = 0  →  no real solutions
solutions = solve_quadratic_equation(1, 0, 1)
print(f"Solutions: {solutions}")  # []

# x² - 4x + 4 = 0  →  (x-2)² = 0  →  x=2 (double root)
solutions = solve_quadratic_equation(1, -4, 4)
print(f"Solutions: {solutions}")  # [2.0]

Where the formula comes from — completing the square

# ax² + bx + c = 0
# x² + (b/a)x = -c/a                    (divide by a, move c)
# x² + (b/a)x + (b/2a)² = -c/a + (b/2a)²  (add same thing to both sides)
# (x + b/2a)² = (b² - 4ac) / 4a²        (left side is now a perfect square)
# x + b/2a = ± √(b² - 4ac) / 2a         (take square root of both sides)
# x = (-b ± √(b² - 4ac)) / 2a           (solve for x)
#
# Every step is algebra — the formula is not magic, it's rearrangement.

Vertex of the parabola

def compute_parabola_vertex(a, b, c):
    """
    The vertex (tip of the parabola) is halfway between the two roots.
    x-coordinate of vertex = -b / (2a)
    y-coordinate = plug that x back in.
    """
    x_of_vertex = -b / (2 * a)
    y_of_vertex = a * x_of_vertex**2 + b * x_of_vertex + c
    return x_of_vertex, y_of_vertex

See also

• Slope and Linear Equations — Rise Over Run
• Pythagorean Theorem — Squares on Sides
• Visual Optimization for Programmers
• Math Foundations — Visual Table of Contents