Free Quadratic Formula Calculator
Enter the coefficients a, b and c to solve a·x² + b·x + c = 0. The calculator returns the roots (real or complex), the discriminant b² − 4ac, and the parabola's vertex, with the working shown.
Solve a·x² + b·x + c = 0. Enter the coefficients a, b and c — the roots, discriminant and vertex update instantly.
x = (−b ± √(b² − 4ac)) / (2a). Discriminant D = b² − 4ac decides the roots: D > 0 two real, D = 0 one repeated, D < 0 two complex. Vertex at x = −b/(2a).
Quick answer
A quadratic equation a·x² + b·x + c = 0 is solved with the quadratic formula x = (−b ± √(b² − 4ac)) / (2a). The discriminant D = b² − 4ac tells you the type of roots: if D > 0 there are two distinct real roots, if D = 0 there is one repeated real root, and if D < 0 there are two complex conjugate roots of the form p ± q·i. This tool computes all of these instantly in your browser, plus the vertex at x = −b/(2a).
Formula & method
Quadratic formula (the roots)
x = (−b ± √(b² − 4ac)) / (2a)
- a — coefficient of x² (must be non-zero)
- b — coefficient of x
- c — constant term
The ± gives the two roots. They are real when the discriminant is ≥ 0 and complex when it is negative.
Discriminant (decides the root type)
D = b² − 4ac
D > 0 → two real roots; D = 0 → one repeated real root; D < 0 → two complex conjugate roots p ± q·i.
Vertex of the parabola y = ax² + bx + c
vertex = ( −b/(2a), c − b²/(4a) )
The x-coordinate is the axis of symmetry; the y-coordinate is the minimum (if a > 0) or maximum (if a < 0).
Examples
- Input
- a = 1, b = −3, c = 2
- Result
- x = 2 and x = 1
- Why
- D = (−3)² − 4·1·2 = 9 − 8 = 1, which is positive. √1 = 1, so x = (3 ± 1)/2, giving (3+1)/2 = 2 and (3−1)/2 = 1. Vertex is at (1.5, −0.25).
- Input
- a = 1, b = −4, c = 4
- Result
- x = 2 (repeated)
- Why
- D = (−4)² − 4·1·4 = 16 − 16 = 0, so there is a single repeated root: x = −b/(2a) = 4/2 = 2. The parabola touches the x-axis at its vertex (2, 0).
- Input
- a = 1, b = 2, c = 5
- Result
- x = −1 + 2i and x = −1 − 2i
- Why
- D = 2² − 4·1·5 = 4 − 20 = −16, which is negative. √(−16) = 4i, so x = (−2 ± 4i)/2 = −1 ± 2i. The roots are complex conjugates; the parabola never crosses the x-axis.
- Input
- a = 2, b = −7, c = 3
- Result
- x = 3 and x = 0.5
- Why
- D = (−7)² − 4·2·3 = 49 − 24 = 25. √25 = 5, so x = (7 ± 5)/(2·2) = (7±5)/4, giving 12/4 = 3 and 2/4 = 0.5.
When to use this tool
- Solving any equation that can be written as ax² + bx + c = 0 when factoring is awkward or impossible.
- Checking homework or exam answers in algebra, and seeing the discriminant and exact steps, not just the roots.
- Finding where a parabola crosses the x-axis, plus its vertex (the maximum or minimum point).
- Modeling problems in physics or engineering — projectile range, break-even points, or any motion under constant acceleration.
Common mistakes
- Forgetting the parentheses around 2a — the entire numerator (−b ± √(b²−4ac)) is divided by 2a, not just the square-root part.
- Sign slips with b. The formula uses −b, so for b = −3 the numerator starts with +3. Squaring also kills the sign: (−4)² = 16, not −16.
- Treating a negative discriminant as 'no solution.' D < 0 means no real roots, but there are still two complex roots p ± q·i.
- Setting a = 0. With a = 0 the equation is linear (bx + c = 0), the quadratic formula divides by zero, and the parabola has no vertex.
Frequently asked questions
What is the quadratic formula?
The quadratic formula is x = (−b ± √(b² − 4ac)) / (2a). It gives the solutions (roots) of any quadratic equation written in the form a·x² + b·x + c = 0, where a is not zero.
What does the discriminant tell me?
The discriminant is D = b² − 4ac, the part under the square root. If D > 0 there are two distinct real roots; if D = 0 there is exactly one repeated real root; and if D < 0 there are two complex conjugate roots and the parabola never touches the x-axis.
Can this calculator handle complex (imaginary) roots?
Yes. When the discriminant is negative, the calculator reports the two complex roots in the form p ± q·i, where p = −b/(2a) and q = √(−D)/(2a). For example, x² + 2x + 5 = 0 returns −1 + 2i and −1 − 2i.
Why must 'a' not be zero?
If a = 0 there is no x² term, so the equation is linear (bx + c = 0), not quadratic. The formula would also divide by 2a = 0, which is undefined. Enter any non-zero value for a; b and c may be zero.
How do I find the vertex of the parabola?
The vertex sits on the axis of symmetry at x = −b/(2a). Substituting back gives the y-coordinate c − b²/(4a). It is the minimum point when a > 0 and the maximum point when a < 0, and the calculator shows both coordinates.
Should I use the quadratic formula or factoring?
Factoring is faster when the roots are simple whole numbers, but the quadratic formula always works — including for messy decimals, irrational roots, and complex roots. When in doubt, the formula never fails.
Sources & references
External references open in a new tab. We are independent and not affiliated with these organizations.
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Provided “as is” for general information only — results may be inaccurate, so verify before you rely on them. No warranty; use at your own risk.
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