A quadratic equation has the form ax² + bx + c = 0. Enter a, b and c – whole numbers, decimals or negative numbers – and the solver gives the roots with every step, plus what they mean for the graph: the vertex, the axis of symmetry and the factored form.
Example: 2x² − 3x − 5 = 0. Here a = 2, b = −3, c = −5. The discriminant is (−3)² − 4 × 2 × (−5) = 9 + 40 = 49; its square root is 7. So x = (3 ± 7) ÷ 4: x₁ = −1 and x₂ = 2.5. Factored: 2(x + 1)(x − 2.5).
Step by step
- Write it as ax² + bx + c = 0. Move every term to one side: 2x² = 3x + 5 becomes 2x² − 3x − 5 = 0.
- Read a, b and c, with their signs. A missing term is 0: x² − 4 = 0 has b = 0.
- Compute the discriminant Δ = b² − 4ac.
- Take its square root and apply x = (−b ± √Δ) ÷ 2a.
- Check one root in the original equation: 2(2.5)² − 3(2.5) − 5 = 12.5 − 7.5 − 5 = 0.
What the discriminant means
| Δ = b² − 4ac | Roots | The parabola |
|---|---|---|
| positive | two real roots | crosses the x-axis twice |
| zero | one double root | touches the x-axis at its vertex |
| negative | two complex roots (conjugates) | never meets the x-axis |
Examples: x² − 5x + 6 = 0 has Δ = 1 and roots 2 and 3; x² − 2x + 1 = 0 has Δ = 0 and the double root 1; x² + 2x + 5 = 0 has Δ = −16 and the roots −1 ± 2i.
Vieta’s formulas: a quick check
For any quadratic, the two roots add up to −b ÷ a and multiply to c ÷ a. In the example, −1 + 2.5 = 1.5 = 3 ÷ 2 and −1 × 2.5 = −2.5 = −5 ÷ 2. If your roots fail this test, recheck the signs.
Other ways to solve
- Factoring: when the roots are whole numbers, find two numbers that multiply to c and add to b (for a = 1): x² − 5x + 6 = (x − 2)(x − 3).
- Completing the square: x² + 6x + 2 = 0 → (x + 3)² = 7 → x = −3 ± √7. The quadratic formula is this method done once for all equations.
- Graphing: the roots are where the parabola crosses the x-axis.
Accurate answers for awkward numbers
When b is very large compared with a and c, the textbook formula subtracts two almost equal numbers and loses the small root to rounding. The solver uses the stable form: it computes q = −(b + sign(b)√Δ) ÷ 2 and then x₁ = q ÷ a, x₂ = c ÷ q. For x² − 100,000,000x + 1 = 0 it finds the small root, 0.00000001, correctly; the plain formula in ordinary computer arithmetic gives 0.0000000075, a quarter too small.
For right triangles, see the Pythagorean theorem calculator. The solver runs on your device, and once zet4u is saved there it works offline too – on the bus or in a library with no Wi-Fi.
Frequently asked questions
What is the quadratic formula?
x = (−b ± √(b² − 4ac)) ÷ 2a, for the equation ax² + bx + c = 0 with a not zero. For 2x² − 3x − 5 = 0: b² − 4ac = 9 + 40 = 49, √49 = 7, so x = (3 ± 7) ÷ 4, which gives x = 2.5 and x = −1.
What does the discriminant tell you?
The number of real solutions. Δ = b² − 4ac. If Δ > 0 there are two different real roots; if Δ = 0, one double root (the parabola touches the x-axis); if Δ < 0, no real roots but two complex ones, −b ÷ 2a ± i√(−Δ) ÷ 2a.
How do I find the vertex of a parabola?
The vertex of y = ax² + bx + c is at x = −b ÷ 2a, and y = c − b² ÷ 4a there. For y = 2x² − 3x − 5 that is (0.75, −6.125). It is the lowest point when a is positive and the highest when a is negative.
What if a is zero?
Then it is not quadratic but linear: bx + c = 0, so x = −c ÷ b. The solver handles that case too, and tells you when an equation has no solution (0 = 5) or every number as a solution (0 = 0).
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