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.
By AdminDrafted with AI writing toolsUpdated Checked examples: 7
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.
Multiply every coefficient by the same nonzero number and the roots stay the same, so clear fractions and make the x² coefficient positive before using the quadratic formula. Doubling x²/2 − 3x/2 + 1 = 0 gives x² − 3x + 2 = 0, with roots 1 and 2.
Changing every sign flips the parabola but keeps the roots. Multiplying −x² + 2x + 1 = 0 by −1 gives x² − 2x − 1 = 0. Both have roots 1 ± √2, but the first parabola peaks at (1, 2) and the second bottoms out at (1, −2).
The discriminant b² − 4ac scales by the square of the multiplier, so its sign, and with it the kind of roots, never changes: doubling x²/2 − 3x/2 + 1 turns Δ = 1/4 into 4 × 1/4 = 1.
The table gives the discriminant of each equation as written, before any scaling:
| a | b | c | Discriminant | Kind of roots |
|---|---|---|---|---|
| 1 | -4 | 1 | 12 | Two distinct real roots |
| 2 | 3 | -5 | 49 | Two distinct real roots |
| 1 | 2 | 5 | −16 | Two complex conjugate roots |
| 1 | -6 | 9 | 0 | One repeated real root |
| 3 | -2 | 1 | −8 | Two complex conjugate roots |
| -1 | 2 | 1 | 8 | Two distinct real roots |
| 1/2 | -3/2 | 1 | 1/4 | Two distinct real roots |
| 1 | 3 | -40 | 169 | Two distinct real roots |
For the complex row 3x² − 2x + 1, Δ = 4 − 12 = −8, so the roots are (1 ± i√2)/3, about 0.3333 ± 0.4714i.
For any equation ax² + bx + c = 0 with a ≠ 0, the solutions are x = (−b ± √(b² − 4ac)) / 2a. For 2x² − 7x + 3 = 0, a = 2, b = −7 and c = 3, so x = (7 ± √25)/4 = (7 ± 5)/4, giving x = 3 or x = 1/2.
The same five steps work on every quadratic:
For whole-number or fractional coefficients, the discriminant 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 |
Check the first row before reaching for the formula: a perfect-square discriminant means the quadratic factors over the integers or fractions. With decimal coefficients from measurements that 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 in standard form w² + 3w − 40 = 0, so a = 1, b = 3 and c = −40.
Both roots solve the equation. Which one to discard depends on the problem, not the algebra.
A ball thrown upward at 14 m/s from a height of 1.5 m, taking g as 9.8 m/s², is at height h = −4.9t² + 14t + 1.5 meters after t seconds. Setting h = 0 gives Δ = 14² − 4 × (−4.9) × 1.5 = 225.4.
The roots simplify to (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: the ball peaks at 11.5 m after 1.43 s.
The formula is what you get by completing the square on 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.
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, then add (6/2)² = 9 to both sides.
That gives (x + 3)² = 16, so x + 3 = ±4 and x = 1 or x = −7. It is 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:
The vertex is (2, −5). Because a = 2 is positive, the parabola opens upward, −5 is its minimum, and the axis of symmetry is the line x = 2.
Setting y = 0 gives 2(x − 2)² = 5, so x = 2 ± √(5/2), also written (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 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 alongside other curves, plot it in the graphing calculator.
David Goldberg's survey of floating-point arithmetic (ACM Computing Surveys, 1991) explains that cancellation and the rearranged formula that avoids it.
This solver uses the rearrangement with 50-digit decimals, so its small root, −1.0000000000000001 × 10⁻⁸, is correct to every digit shown.
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: Independent calculation
Checked against: Δ = 9 + 40 = 49, (−3 ± 7)/4
Checked against: Independent calculation
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.
This calculator includes 7 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.