Quadratic Equation Solver
Enter a, b, c of ax²+bx+c=0 to get roots x₁, x₂, the discriminant Δ, the vertex (h,k) and complex roots, with a parabola chart.
Input Data
Results
Parabola y = 1x² + -3x + 2
Chart notes:The parabola y = ax²+bx+c has axis of symmetry x = h; its opening is set by the sign of a (a>0 up, a<0 down), and the vertex is the extremum.
At a glance:A quadratic is ax²+bx+c=0 (a≠0). The discriminant Δ=b²−4ac gives Δ>0 two distinct real roots, Δ=0 a double root, Δ<0 complex conjugates −b/2a ± i·√|Δ|/2a. The vertex (h,k) with h=−b/2a and k=−Δ/4a is the parabola's highest/lowest point.
Formula
Discriminant: Δ = b² − 4ac.
Roots: x = (−b ± √Δ) / (2a).
Vertex: h = −b / (2a), k = f(h) = −Δ / (4a).
Δ<0 (no real root): x = −b/(2a) ± i·√|Δ|/(2a).
$$\Delta = b^{2} - 4ac$$$$x = \frac{-b \pm \sqrt{\Delta}}{2a}$$$$h = -\frac{b}{2a},\quad k = -\frac{\Delta}{4a}$$How to Use
- Enter the quadratic coefficient a (a must not be 0).
- Enter the linear coefficient b and the constant c.
- The tool instantly returns roots x₁, x₂, the discriminant Δ and the vertex (h,k), and lists the complex conjugate roots in the note when Δ<0.
- Use the case studies below to check your own working or verify an answer.
Common quadratic examples (roots and vertex)
| Equation ax²+bx+c=0 | Δ | x₁ | x₂ | Vertex (h, k) |
|---|---|---|---|---|
| x² − 3x + 2 = 0 | 1 | 2 | 1 | (1.5, −0.25) |
| x² + 2x + 1 = 0 | 0 | −1 | −1 | (−1, 0) |
| x² + x + 1 = 0 | −3 | −0.5 | ±0.8660i | (−0.5, 0.75) |
| 2x² − 4x − 6 = 0 | 64 | 3 | −1 | (1, −8) |
| x² − 4 = 0 | 16 | 2 | −2 | (0, −4) |
| 3x² + 6x = 0 | 36 | 0 | −2 | (−1, −3) |
When Δ<0, x₂ shows the imaginary part as ±bi; otherwise both are real roots.
Case Studies
x² − 3x + 2 = 0 (two real roots)
Δ = (−3)² − 4·1·2 = 9 − 8 = 1 > 0.
x = (3 ± √1) / 2 = (3 ± 1) / 2 → x₁ = 2, x₂ = 1.
Vertex h = 3/2 = 1.5, k = −Δ/4a = −1/4 = −0.25; upward-opening parabola, minimum at (1.5, −0.25).
x² + 2x + 1 = 0 (double root)
Δ = 2² − 4·1·1 = 4 − 4 = 0.
x = −2 / 2 = −1, a repeated (double) root.
The vertex is (−1, 0); the parabola touches the x-axis.
x² + x + 1 = 0 (complex conjugates)
Δ = 1² − 4·1·1 = 1 − 4 = −3 < 0, no real root.
Real part = −b/2a = −0.5; imaginary part = √|Δ|/2a = √3/2 ≈ 0.8660.
Complex conjugate roots: x = −0.5 ± 0.8660 i.
2x² − 4x − 6 = 0 (a ≠ 1)
Δ = (−4)² − 4·2·(−6) = 16 + 48 = 64.
x = (4 ± √64) / 4 = (4 ± 8) / 4 → x₁ = 3, x₂ = −1.
Vertex h = 4/4 = 1, k = −64/8 = −8.
Projectile: height h(t) = −5t² + 20t + 2 = 0
Take a=−5, b=20, c=2 and solve for landing time (h=0).
Δ = 20² − 4·(−5)·2 = 400 + 40 = 440; √440 ≈ 20.976.
t = (−20 ± 20.976) / (−10) → the positive root t ≈ 4.098 s; the vertex at t = 2 s is the peak.
FAQ
What if a = 0?
The equation becomes linear bx + c = 0, which this tool does not solve (a≠0 required). Use a linear-equation solver, or set a very small value as an approximation.
What does the discriminant Δ mean?
Δ = b² − 4ac decides the root type: Δ>0 two distinct real roots, Δ=0 a double root, Δ<0 no real root (complex conjugates). It comes from the radicand in the quadratic formula.
What is the meaning of complex roots?
With real coefficients and Δ<0 the roots are complex conjugates. Physically this often means no real intersection — for example the parabola stays above the x-axis and never lands.
How are the vertex and the roots related?
The vertex x-coordinate h = −b/2a is exactly the midpoint of the two real roots. If the roots are r₁, r₂ then h = (r₁+r₂)/2, and you can recover the factorization a(x−r₁)(x−r₂)=0.
Why compute the vertex?
The vertex (h,k) is the highest or lowest point of y = ax²+bx+c and gives the extremum. It is used in optimization (max profit, min cost) and in projectile peak/range analysis.
How is floating-point error handled?
Results are rounded to 4 decimal places. When Δ is near 0 it may be misclassified by rounding; treat |Δ| within ±0.0001 as a double root.
Vertex and parabola direction?
a>0 opens upward with a minimum vertex; a<0 opens downward with a maximum vertex. Both share the vertical axis of symmetry x = h.
How to recover the equation from the roots?
Given roots r₁, r₂, the equation is a(x−r₁)(x−r₂)=0; expand and compare coefficients to recover a, b, c.
How does the chart relate to the roots?
The plotted parabola y = ax²+bx+c has x-intercepts equal to the real roots; its lowest/highest point is the vertex (h,k). When Δ<0 the parabola does not cross the x-axis.
Related Tools
References
Content review: Calculatorism Science Team. Quadratic formula, the three discriminant cases and the vertex formula have been verified. Results are for reference only.