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Last updated: March 21, 2026

Quadratic Equation Solver

How do you solve a quadratic equation?

With the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a. The discriminant Δ = b² - 4ac decides the outcome: Δ>0 two real solutions, Δ=0 one repeated solution, Δ<0 no real solution.

Calculator

x²

Quadratic Equation Solver

Solve ax² + bx + c = 0 with formula and steps

ax² + bx + c = 0
a ≠ 0

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Quadratic Equations: Theory and Solution

A quadratic equation (or second-degree equation) is an algebraic equation in which the highest degree of the unknown is 2. It is one of the pillars of algebra and appears in countless applications: from physics to trajectory calculations, from economics to engineering.

Standard Form

Every quadratic equation can be written in standard form:

ax² + bx + c = 0

where a, b, and c are real coefficients and a ≠ 0 (if a were zero, the equation would be linear).

The Discriminant

The key to solving a quadratic equation is the discriminant (denoted by the Greek letter delta):

Δ = b² – 4ac

The value of the discriminant determines the number and nature of the solutions:

Value of Δ Real Solutions Description
Δ > 0 2 distinct The parabola intersects the x-axis at two points
Δ = 0 1 (repeated) The parabola is tangent to the x-axis
Δ < 0 0 (2 complex) The parabola does not intersect the x-axis

The Quadratic Formula

The solutions (roots) of the equation are obtained using the quadratic formula:

x = (–b ± √Δ) / (2a)

which, expanding the discriminant, becomes:

x = (–b ± √(b² – 4ac)) / (2a)

Examples for Each Case

Case 1: Δ > 0 — Two Distinct Real Solutions

Solve x² – 5x + 6 = 0 (a=1, b=–5, c=6)

  • Δ = (–5)² – 4(1)(6) = 25 – 24 = 1
  • x₁ = (5 + 1) / 2 = 3
  • x₂ = (5 – 1) / 2 = 2

Case 2: Δ = 0 — Repeated Solution

Solve x² – 6x + 9 = 0 (a=1, b=–6, c=9)

  • Δ = (–6)² – 4(1)(9) = 36 – 36 = 0
  • x = 6 / 2 = 3 (repeated solution)

Case 3: Δ < 0 — No Real Solutions

Solve x² + 2x + 5 = 0 (a=1, b=2, c=5)

  • Δ = (2)² – 4(1)(5) = 4 – 20 = –16
  • No real solutions exist
  • Complex solutions: x = (–2 ± 4i) / 2 → x₁ = –1 + 2i, x₂ = –1 – 2i

Vertex of the Parabola

The equation ax² + bx + c = 0 is associated with the parabola y = ax² + bx + c. The vertex of the parabola has coordinates:

V = (–b/(2a), –Δ/(4a))

The vertex represents the minimum point (if a > 0) or the maximum point (if a < 0) of the parabola. For the equation x² – 5x + 6 = 0:

  • xᵥ = 5 / 2 = 2.5
  • yᵥ = –1 / 4 = –0.25
  • Vertex: V(2.5, –0.25)

Relationships Between Roots and Coefficients (Vieta's Formulas)

If x₁ and x₂ are the roots of the equation, the following relationships hold:

  • x₁ + x₂ = –b/a (sum of roots)
  • x₁ × x₂ = c/a (product of roots)

These formulas allow you to quickly verify the solutions found and are useful for constructing equations from desired roots.

Special Cases

Type Condition Quick Method
Pure b = 0 → ax² + c = 0 x = ±√(–c/a)
Incomplete c = 0 → ax² + bx = 0 x(ax + b) = 0 → x=0, x=–b/a
Trivial b = 0 and c = 0 x = 0 (repeated)

How to Use the Calculator

Enter the coefficients a, b, and c of the equation in standard form. The solver will calculate the discriminant, the solutions (real or complex), the factored form (if Δ ≥ 0), the coordinates of the parabola's vertex, the axis of symmetry, and the concavity. Decimal and negative values are accepted for the coefficients.

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