Quadratic Formula Calculator

Solves ax² + bx + c = 0 with the quadratic formula and shows exact roots such as (−3 ± √17)/4 with decimal values, plus the discriminant, vertex, graph and worked steps.

How to use

1. Type the equation into "Equation", for example x^2 - 5x + 6 = 0, or enter the coefficients in a, b and c. Fractions and decimals work.
2. Read the roots under "Solution", with the discriminant, the kind of roots, and their sum and product.
3. Under "Steps", choose "Quadratic formula", "Factoring" or "Completing the square" to see the working.
4. The "Graph" marks the x-intercepts and vertex; "Properties of the graph and roots" lists the vertex, axis of symmetry, y-intercept and vertex form.

How it works

• x = (−b ± √(b² − 4ac)) ÷ 2a, with discriminant D = b² − 4ac. D > 0 gives two real roots, D = 0 one repeated root, D < 0 two complex roots.
• Roots are calculated exactly and fully simplified; complex roots use i.
• If a = 0, the equation is solved as a linear one. Equations of degree 3 or more, or with two variables, are not solved.

Questions and answers

What is the quadratic formula?

For ax² + bx + c = 0, x = (−b ± √(b² − 4ac)) ÷ 2a. For x² − 5x + 6 = 0 this gives x = (5 ± 1) ÷ 2, so x = 2 or 3.

What does the discriminant tell you?

The number of real roots: two if b² − 4ac is positive, one repeated root if it is 0, and none (two complex roots) if it is negative.

Is there a shortcut when b is even?

Yes. With b = 2b', x = (−b' ± √(b'² − ac)) ÷ a. When b is even, the "Quadratic formula" steps include it under "Shortcut formula for an even b".

How do I find the vertex of a parabola?

Its x-coordinate is −b ÷ 2a; put that into the equation for the y-coordinate. For y = x² − 4x + 3 the vertex is (2, −1).