Solve ax² + bx + c = 0 with the quadratic formula: exact roots as surds or complex numbers, the discriminant, vertex and axis of symmetry, and a graph.
The discriminant is positive (12), so the parabola crosses the x-axis twice: x = 2 ± √3. It opens upward with a minimum of −3 at x = 2.
A quadratic equation ax² + bx + c = 0, with a ≠ 0, has the roots x = (−b ± √(b² − 4ac)) / 2a. The discriminant Δ = b² − 4ac decides their kind: two real roots when Δ > 0, one repeated root when Δ = 0, and a pair of complex conjugates when Δ < 0. The vertex of the parabola y = ax² + bx + c sits on its axis of symmetry, x = −b/2a.
Projectile heights, areas with one unknown side and break-even points all lead to quadratics. The default, x² − 4x + 1 = 0, has Δ = 12, so its roots are x = 2 ± √3, about 0.2679 and 3.7321, and its vertex is (2, −3).
Roots are given exactly, as fractions, simplified surds or complex numbers, and as decimals to 10 places. The decimals use the cancellation-free form from Numerical Recipes, so a tiny root next to a large one keeps its accuracy. With a = 0 the equation is linear, not quadratic.
The formula is what you get by completing the square on the general equation ax² + bx + c = 0 once, with letters instead of numbers:
Step 4 is where the discriminant enters. The right side of step 3 has a positive denominator, 4a², so the sign of b² − 4ac alone decides whether a real square root exists.
For whole-number or fractional coefficients, Δ = b² − 4ac sorts every quadratic into one of four cases.
| Discriminant Δ | Roots | Graph of y = ax² + bx + c | Example | Δ | Roots of the example |
|---|---|---|---|---|---|
| Positive perfect square | Two rational roots | Crosses the x-axis twice | 3x² − 5x − 2 | 49 | −1/3 and 2 |
| Positive, not a perfect square | Two irrational roots, a ± √ pair | Crosses the x-axis twice | x² + 6x + 4 | 20 | −3 ± √5 |
| Zero | One repeated root | Touches the x-axis at the vertex | 4x² − 12x + 9 | 0 | 3/2 |
| Negative | Two complex conjugate roots | Stays on one side of the x-axis | x² − 4x + 13 | −36 | 2 ± 3i |
The first row is the one worth checking before reaching for the formula: a perfect-square discriminant means the quadratic factors over the integers or fractions. With decimal coefficients that come from measurements, the perfect-square case is rare and the formula is the practical route.
A rectangular garden is 3 m longer than it is wide and covers 40 m². Calling the width w, the area gives w(w + 3) = 40, and moving everything to one side puts it in standard form: w² + 3w − 40 = 0, so a = 1, b = 3 and c = −40.
The solver returns x₁ = −8 and x₂ = 5 for a = 1, b = 3, c = −40. It reports both roots because both solve the equation; which one to discard depends on the problem, not the algebra.
A projectile gives a second case where one root is thrown away. A ball thrown upward at 14 m/s from a height of 1.5 m, taking g as 9.8 m/s², has height h = −4.9t² + 14t + 1.5 metres after t seconds. Setting h = 0 gives Δ = 14² − 4 × (−4.9) × 1.5 = 225.4, and the solver shows the roots exactly as (10 ± √115)/7, about −0.1034 and 2.9605. The negative time lies before the throw, so the ball lands after 2.96 s. The vertex, at t = 1.4286 and h = 11.5, is the top of the flight: 11.5 m after 1.43 s.
When the discriminant is a perfect square, factoring is often quicker than the formula. The ac method handles a leading coefficient other than 1. For 6x² + x − 2 = 0:
Completing the square works on any quadratic and is quickest when a = 1 and b is even. For x² + 6x − 7 = 0, move the constant to get x² + 6x = 7, add (6/2)² = 9 to both sides to get (x + 3)² = 16, and take roots: x + 3 = ±4, so x = 1 or x = −7. It is the same procedure as the derivation above, applied to numbers.
Completing the square without setting the expression to zero rewrites y = ax² + bx + c in vertex form, y = a(x − h)² + k, where h = −b/2a and k = c − b²/4a. For y = 2x² − 8x + 3:
Vertex form reads off three facts at once. The vertex is (2, −5); a = 2 is positive, so the parabola opens upward and −5 is its minimum; and the axis of symmetry is the line x = 2. The roots follow by setting y = 0: 2(x − 2)² = 5, so x = 2 ± √(5/2), which the solver writes as (4 ± √10)/2, about 0.4189 and 3.5811. Both sit 1.5811 either side of the axis, as every pair of real roots does.
Two quick checks catch most arithmetic slips. The roots of any quadratic add up to −b/a and multiply to c/a. Here 0.4189 + 3.5811 = 4 = 8/2 and 0.4189 × 3.5811 = 1.5 = 3/2. To see the parabola itself alongside other curves, plot it in the graphing calculator.
Complex roots, such as 2 ± 3i above, can be carried into further arithmetic with the complex number calculator. For cubic to sixth-degree equations, the polynomial root finder returns every real and complex root.
Checked against: Python decimal (60 digits): 2 ± Decimal(3).sqrt()
Checked against: Δ = 9 + 40 = 49, (−3 ± 7)/4 (Python fractions)
Checked against: Python cmath: (-2 ± cmath.sqrt(-16)) / 2 = -1 ± 2j
Checked against: (x − 3)² expanded by hand
x = (−b ± √(b² − 4ac)) / 2a gives both solutions of ax² + bx + c = 0. For 2x² + 3x − 5 = 0, a = 2, b = 3 and c = −5, so x = (−3 ± √49)/4 = (−3 ± 7)/4, giving x = 1 and x = −5/2. The formula comes from completing the square and works for every quadratic with a ≠ 0.
The discriminant b² − 4ac counts the real roots. Positive means two different real roots (x² − 4x + 1 has Δ = 12); zero means one repeated root where the parabola touches the x-axis (x² − 6x + 9 has Δ = 0 and root 3); negative means two complex roots (x² + 2x + 5 has Δ = −16). With whole-number coefficients, a perfect-square discriminant such as 49 means the roots are rational.
The vertex x-coordinate is −b/2a; substituting it back gives the y-coordinate, which equals c − b²/4a. For y = x² − 4x + 1 the vertex is at x = 4/2 = 2 and y = 1 − 16/4 = −3. If a > 0 the parabola opens upward and the vertex is its minimum; if a < 0 it opens downward and the vertex is its maximum.
When b² − 4ac is negative the square root in the formula is imaginary, so the roots take the form p ± qi, where i = √−1. For x² + 2x + 5 = 0, Δ = 4 − 20 = −16 and √−16 = 4i, so x = (−2 ± 4i)/2 = −1 ± 2i. The parabola never crosses the x-axis, and the real part, −1, is the x-coordinate of its vertex.
Yes, when the roots are rational. x² − 6x + 9 factors as (x − 3)², so x = 3 twice, and 2x² + 3x − 5 factors as (2x + 5)(x − 1), giving −5/2 and 1. x² − 4x + 1 does not factor over whole numbers because its discriminant, 12, is not a perfect square, so the formula's answer 2 ± √3 is needed.
Accuracy depends on your inputs and the method's assumptions. Decimal arithmetic uses 50 significant digits, but estimates, numerical methods and source data can be less precise; the displayed rounding does not remove those limits. It is checked against 7 worked examples whose answers come from independent sources; for example, “x² − 4x + 1 (irrational roots)” is checked against Python decimal (60 digits): 2 ± Decimal(3).sqrt().
Wolfram MathWorld — Quadratic Equation; Press et al., Numerical Recipes (3rd ed.) §5.6 — cancellation-free quadratic roots.
7 worked examples with independently sourced answers ship with this calculator. They run in the test suite; you can run them here too.
All real and complex roots of a polynomial up to degree 6, such as a cubic or quartic equation, with repeated roots found exactly and 30-digit accuracy.
Add, subtract, multiply and divide complex numbers, raise them to powers, find nth roots, and convert a + bi to polar form, with an Argand diagram.
Solve 2 to 5 simultaneous linear equations exactly by Gauss–Jordan elimination, with every row operation shown, and detect no-solution or infinite cases.
Raise a number to any exponent, including fractional and negative ones, or take its square, cube or nth root, with exact simplified radicals such as 5√2.
The logarithm of a number to any base, including ln and log₁₀, with change-of-base steps, or the exponent x that solves bˣ = y, exact when rational.
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