Solve ax² + bx + c = 0 using the quadratic formula. Finds real and complex roots, discriminant, vertex, and axis of symmetry.
Enter coefficients for ax² + bx + c = 0
| Formula | Expression |
|---|---|
| Roots | x = (−b ± √(b²−4ac)) / 2a |
| Discriminant | Δ = b² − 4ac |
| Vertex x (h) | h = −b / 2a |
| Vertex y (k) | k = c − b²/4a |
| Sum of roots | x₁ + x₂ = −b/a |
| Product of roots | x₁ · x₂ = c/a |
| Property | Value |
|---|---|
| x₁ | 3 |
| x₂ | 2 |
| Discriminant (Δ) | 1 |
| Vertex (h, k) | (2.5, -0.25) |
| Axis of symmetry | x = 2.5 |
| y-intercept | (0, 6) |
| Parabola opens | Upward ↑ (minimum at vertex) |
| Δ | Roots | Graph |
|---|---|---|
| Δ > 0 | Two distinct real roots | Crosses x-axis twice |
| Δ = 0 | One repeated real root | Touches x-axis once |
| Δ < 0 | Two complex conjugate roots | Never crosses x-axis |
x² + 2x + 5 = 0 has two solutions, yet its parabola never touches the x-axis: the lowest point is (−1, 4), four units above it. The calculator reports the roots as −1 + 2i and −1 − 2i, and that −1 is exactly the x-coordinate of the vertex. This is no coincidence, and seeing roots, vertex and discriminant side by side makes the pattern easy to spot.
For ax² + bx + c = 0 with a ≠ 0, the roots are x = (−b ± √(b² − 4ac)) ÷ 2a. The quantity under the root, Δ = b² − 4ac, is the discriminant. Positive gives two real roots, zero gives one repeated root, and negative gives a complex pair written (−b ÷ 2a) ± (√(−Δ) ÷ 2a)i. The page treats any Δ within 1e-10 of zero as zero, to absorb floating-point noise.
The vertex sits at h = −b ÷ 2a and k = c − b² ÷ 4a, and the axis of symmetry is the line x = h. The two roots are mirror images across that line, so their average is h, which is why the real part of a complex pair equals h. Two quick checks apply to any answer: the roots add to −b ÷ a and multiply to c ÷ a. Outputs are rounded to 8 decimals.
Worked example: 2x² − 3x − 5 = 0, so a = 2, b = −3, c = −5. Δ = (−3)² − 4 × 2 × (−5) = 9 + 40 = 49 and √49 = 7. Then x₁ = (3 + 7) ÷ 4 = 2.5 and x₂ = (3 − 7) ÷ 4 = −1. The check passes: 2.5 + (−1) = 1.5 = −b ÷ a, and 2.5 × (−1) = −2.5 = c ÷ a. The vertex is h = 3 ÷ 4 = 0.75 and k = −5 − 9 ÷ 8 = −6.125. Because a is positive, the parabola opens upward with its minimum at (0.75, −6.125) and crosses the y-axis at (0, −5).
Move everything to one side first. 3x² = 12 − x becomes 3x² + x − 12 = 0, so a = 3, b = 1, c = −12, which gives Δ = 145.
Use 0 for that coefficient. For x² − 9 = 0, set b = 0 and c = −9, and the roots come out as 3 and −3.
The page divides the imaginary part by 2a, so when a is negative the sign sits inside the number and a line can read 0 + -1i. The pair is still correct, just awkwardly written, and the second root reads 0 − -1i.
No. x² − 2 = 0 returns 1.41421356 and −1.41421356, rounded to 8 decimals. The exact values are ±√2.
Then there is no x² term and the equation is linear. Solve bx + c = 0 directly as x = −c ÷ b.
Set height to zero in h = −4.9t² + 20t + 1.5, so a = −4.9, b = 20, c = 1.5. Δ = 429.4 and the roots are t = −0.07367031 and t = 4.15530296. Discard the negative time; the ball lands after about 4.16 seconds.