Quadratic equation solver — roots, vertex and discriminant
Find the values of x that solve a quadratic equation, then check the discriminant, turning point and factored form where the roots are real.
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When the discriminant is negative the roots are complex conjugates — the parabola never crosses the x-axis.
How to use
- Put the equation in standard form. Move all terms to one side so it reads ax² + bx + c = 0. Include the minus signs when entering coefficients.
- Enter a, b and c. For x² − 5x + 6 = 0, enter 1, −5 and 6. The solutions are x = 2 and x = 3.
- Check the result. That example factors as (x − 3)(x − 2) = 0 and has a vertex at (2.5, −0.25). Displayed results may be rounded.
Questions
What if a coefficient is missing?
Enter 0 for a missing x term or constant: x² + 1 = 0 uses a = 1, b = 0 and c = 1. If a is 0, the equation is no longer quadratic.
What does the discriminant tell me?
The discriminant is b² − 4ac. Positive gives two distinct real roots; zero gives one repeated real root; negative gives a complex conjugate pair. Real roots correspond to points where the graph meets the x-axis.
What are complex roots?
They contain i, where i² = −1. For x² + 1 = 0 the roots are +i and −i. They solve the equation even though its real-valued graph does not meet the x-axis.
How do I find the vertex?
Its x-coordinate is −b ÷ (2a); substitute that into the equation’s left-hand side to find y. It is the minimum when a is positive and the maximum when a is negative.
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