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Quadratic Equation Solver

Use this free quadratic equation solver to find real and complex roots instantly. Whether you want to solve by factoring or need vertex form coordinates, get clear steps for any second-degree polynomial.

What is a Quadratic Equation?

A quadratic equation is a second-order polynomial with the standard form ax² + bx + c = 0, where x represents the unknown variable and a, b, and c are fixed coefficients (with a ≠ 0).

Graphing a quadratic function creates a smooth U-shaped curve called a parabola. The points where the parabola crosses the horizontal x-axis represent the equation's real roots or solution values.

The Quadratic Formula

When you cannot solve a quadratic equation by factoring, apply the algebraic quadratic formula:

x =
-b ± √(b² - 4ac)
2a

The expression underneath the square root, b² - 4ac, is the discriminant (Δ). It reveals whether solutions are real or complex before completing the arithmetic.

Step-by-Step Calculation Logic

1. Arrange in Standard Form: Move all terms to one side so the equation reads ax² + bx + c = 0. Identify values for coefficients a, b, and c.

2. Evaluate the Discriminant: Compute Δ = b² - 4ac to check root types:

  • Δ > 0: Yields two distinct real roots (crosses x-axis twice).
  • Δ = 0: Yields one repeated real root (touches x-axis at its vertex).
  • Δ < 0: Yields two complex/imaginary roots (never touches the x-axis).

3. Apply the Formula: Substitute a, b, and Δ into (-b ± √Δ) / 2a. Split the ± symbol into two operations to resolve both values of x.

4. Worked Example: For 2x² - 4x - 6 = 0 (a = 2, b = -4, c = -6):

Δ = (-4)² - 4(2)(-6) = 16 + 48 = 64

x = (4 ± √64) / (2 · 2) = (4 ± 8) / 4

x₁ = 12 / 4 = 3, x₂ = -4 / 4 = -1

How to Find the Vertex of a Parabola

Every parabola has a central peak or trough called the vertex (h, k). Finding the vertex helps convert standard polynomials into vertex form: y = a(x - h)² + k.

  • Axis of Symmetry (h): Calculate h = -b / (2a). This vertical line splits the parabola into equal halves.
  • Vertex Height (k): Plug h back into your equation: k = a(h)² + b(h) + c.

Frequently Asked Questions

What is the quadratic formula?

The quadratic formula is x = (-b ± √(b² - 4ac)) / (2a). It provides exact solutions for unknown variable x in any standard quadratic polynomial ax² + bx + c = 0.

How do you solve a quadratic equation step by step?

First, balance your polynomial into standard form ax² + bx + c = 0. Next, isolate coefficients a, b, and c. Compute the discriminant (b² - 4ac), plug your values into the quadratic formula, and solve for both positive and negative branches.

What does the discriminant tell you in a quadratic equation?

The discriminant (b² - 4ac) indicates your root types. If it is positive, you get two real roots; if zero, one real repeated root; and if negative, two complex roots containing imaginary units (i).

Can a quadratic equation have no real solution?

Yes. When the discriminant is negative (b² - 4ac < 0), the parabola never touches the x-axis on a standard cartesian graph. The solutions are pairs of complex conjugate roots with imaginary numbers.

Quadratic Equation Solver

Presets:••
x² - 5x + 6 = 0
Calculated Solutions (Roots)Opens ∪
First Root (x₁)3
Second Root (x₂)2
Nature of Roots: Two Real Roots (Distinct)Δ = 1
Key Parabola Properties
Vertex Point (h, k)(2.50, -0.25)Extremum / Peak
Axis of Symmetryx = 2.50Center vertical line
Y-Intercept(0, 6)Crosses vertical axis
Discriminant (b² - 4ac)1Root test value
Factored Form:(x - 3.00)(x - 2.00) = 0
Vertex Form:(x - 2.50)² - 0.25 = 0
📐 Step-by-Step Quadratic Formula Solution

1. Identify coefficients: a = 1, b = -5, c = 6

2. Calculate Discriminant: Δ = b² - 4ac = (-5)² - 4(1)(6) = 25 - (24) = 1

3. Apply Quadratic Formula: x = [-b ± √Δ] / 2a = [-(-5) ± √1] / 2(1)

4. Root 1: x₁ = (5 + 1.0000) / 2 = 3

5. Root 2: x₂ = (5 - 1.0000) / 2 = 2

* Roots of a quadratic equation ax² + bx + c = 0 are calculated via the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a. A positive discriminant yields two real roots, zero produces one repeated root, and a negative discriminant results in complex conjugates.

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