How it works
What the quadratic formula does
A quadratic equation has the general shape ax² + bx + c = 0 with a ≠ 0. The quadratic formula (known as Bhaskara in Brazilian and Portuguese textbooks, after the 12th-century Indian mathematician Bhāskara II) gives every possible value of x that satisfies the equation, in one step.
It works whether the coefficients are integers, decimals or letters — which is why it beats factoring as a general-purpose tool. Our widget accepts a, b and c and returns the discriminant, both roots and the factored form.
The formula, step by step
- Step 1 — write the equation as ax² + bx + c = 0 with 0 on the right.
- Step 2 — read off the coefficients a, b, c.
- Step 3 — compute the discriminant Δ = b² − 4ac.
- Step 4 — if Δ < 0 the roots are complex: x = (−b ± i√|Δ|) / 2a.
- Step 5 — else apply x = (−b ± √Δ) / 2a to get the two real roots.
Three worked examples
1. Two distinct real roots — x² − 5x + 6 = 0
a = 1, b = −5, c = 6.
Δ = (−5)² − 4 × 1 × 6 = 25 − 24 = 1.
x = (5 ± √1) / 2 → x₁ = 3, x₂ = 2.
Cross-check by factoring: (x − 2)(x − 3) = 0. ✓
2. Repeated root — x² − 6x + 9 = 0
a = 1, b = −6, c = 9.
Δ = 36 − 36 = 0 → single real root.
x = 6 / 2 = 3 (multiplicity 2). The parabola touches the x-axis once.
3. Complex roots — x² + 2x + 5 = 0
a = 1, b = 2, c = 5.
Δ = 4 − 20 = −16 → no real roots.
x = (−2 ± √−16) / 2 = −1 ± 2i. The parabola sits entirely above the x-axis.
What the discriminant tells you
- Δ > 0 — two distinct real roots; the parabola crosses the x-axis twice.
- Δ = 0 — one real root (double); the parabola is tangent to the x-axis.
- Δ < 0 — two complex conjugate roots; the parabola doesn't touch the x-axis.
- Vertex — the x-coordinate of the minimum/maximum is x = −b/2a, regardless of Δ.
- Sum and product of roots (Vieta's formulas): x₁ + x₂ = −b/a and x₁ × x₂ = c/a. Handy for cross-checks.
When to factor vs when to use the formula
Factoring is faster when a, b and c are small integers and the roots are obvious (like x² − 7x + 12). The formula wins when coefficients are ugly, the roots involve radicals, or you need a programmatic answer.
Completing the square sits between the two and is how Bhaskara's formula is actually derived — useful to know for A-level exam proofs.
Real-world uses
- Projectile motion — the height h(t) = h₀ + v₀t − ½gt² is quadratic in t; solving h = 0 tells you when the projectile lands.
- Profit maximisation — revenue − cost is often a downward-opening parabola; the vertex gives the profit-maximising price.
- Area problems — "a rectangle has perimeter 40 m and area 96 m²" becomes x(20 − x) = 96, a quadratic in x.
- Physics / engineering — resonant frequency, LC circuits, RLC damping — all quadratic in nature.
Works well with
- **Pythagoras calculadora** — sides of right triangles often lead to quadratic equations.
- **Exponentiation calculadora** — powers appear throughout quadratic derivations.
- **GCD calculadora** — useful when simplifying roots in surd form.
- **Rule of three** — quick proportional sanity check on answers.
Deriving the formula by completing the square
Examiners at GCSE and A-level love to ask for the proof, not just the answer. The trick is called completing the square and it is the reason Bhaskara's formula exists at all. Start from the general quadratic and work it into a perfect-square shape.
ax² + bx + c = 0. Divide through by a (valid because a ≠ 0): x² + (b/a)x + c/a = 0. Move c/a across: x² + (b/a)x = −c/a.
Add (b/2a)² to both sides so the left becomes a perfect square: x² + (b/a)x + (b/2a)² = (b/2a)² − c/a. The left is now (x + b/2a)². Tidy the right: (b² − 4ac) / 4a².
Take square roots: x + b/2a = ±√(b² − 4ac) / 2a. Move b/2a across and you land on x = (−b ± √(b² − 4ac)) / 2a — the formula we started with. The group inside the radical, b² − 4ac, is what we later call the discriminant Δ.
Vieta's formulas — the quick sanity check
Before a calculator existed, students used Vieta's (Girard's) relations to check answers in seconds: sum of roots = −b/a and product of roots = c/a. They come straight out of expanding a(x − x₁)(x − x₂) = 0, and they save time in exam conditions.
Quick example
Take 2x² − 3x − 5 = 0. The formula gives x₁ = 5/2 and x₂ = −1. Check: sum = 5/2 + (−1) = 3/2 = −b/a = 3/2 ✓. Product = 5/2 × (−1) = −5/2 = c/a = −5/2 ✓. If either check fails, you made an arithmetic slip in the formula — re-read your working before writing more.
Building an equation from known roots
Given roots 7 and −2, the monic quadratic with those roots is x² − (sum) x + (product) = x² − 5x − 14 = 0. This trick is useful whenever a problem hands you the roots and asks for the equation, which appears in nearly every GCSE and A-level non-calculator paper.
The shape of the parabola — vertex, axis and intercepts
Every quadratic is a parabola. The sign of a tells you the opening direction: a > 0 opens upwards, a < 0 opens downwards. The axis of symmetry is the vertical line x = −b/2a, which is also the x-coordinate of the vertex. Substitute that back into the equation to get the minimum (or maximum) value, y = c − b²/4a.
- y-intercept — when x = 0, y = c. Always the easiest point to plot.
- x-intercepts — the real roots from Bhaskara. If Δ < 0 there are none; the parabola sits wholly above or below the x-axis.
- Vertex — (−b/2a, c − b²/4a). Exam papers routinely ask for the "turning point", which is exactly this.
- Concavity and stretch — the larger |a|, the steeper the sides. a = 1 is the baseline shape; a = 2 is twice as narrow; a = 0.5 is twice as wide.
Two exam-style walkthroughs
These match the style of recent GCSE and A-level papers. Work each one through with pen and paper before checking the answer.
Problem 1 — kinematics (AS-level mechanics)
A ball is thrown upwards from a 2 m cliff with initial velocity 12 m/s. Using g = 9.8 m/s², its height above the ground is h(t) = 2 + 12t − 4.9t². When does it hit the ground? Setting h(t) = 0 gives −4.9t² + 12t + 2 = 0, or 4.9t² − 12t − 2 = 0. Discriminant Δ = 144 + 39.2 = 183.2. √Δ ≈ 13.535. t = (12 + 13.535) / 9.8 ≈ 2.60 s (the negative root is rejected because time cannot be negative).
Problem 2 — optimisation (GCSE higher tier)
A rectangular enclosure uses 60 m of fencing, with one side formed by an existing wall. What width x maximises the area? The two fenced sides parallel to the wall sum to 60 − 2x, so area A(x) = x(60 − 2x) = −2x² + 60x. This is a downward-opening parabola; the vertex is at x = −b/2a = −60 / (2 × −2) = 15 m. Maximum area = 15 × 30 = 450 m². Bhaskara is not needed for the answer, but the same formula gives the roots 0 and 30 — the widths at which the area vanishes.
Common mistakes with the quadratic formula
Most wrong answers are not about the formula itself — they are about the numbers you feed into it. Watch for these five.
- Sign slip on b. If b = −7, plugging in gives −(−7) = +7, not −7. Miss this and every subsequent step is wrong.
- Squaring a negative. (−7)² = 49, not −49. Most calculators follow the standard rule if you bracket the negative, but a bare −7² on a non-scientific key sequence returns −49.
- Dividing only one term by 2a. The whole numerator (−b ± √Δ) is divided by 2a. Students often divide only −b and leave the radical untouched.
- Forgetting the ± gives two roots. "Just one x" is an incomplete answer except when Δ = 0.
- Ignoring non-real roots when physics demands them. In pure maths, Δ < 0 is an accepted answer (complex roots). In kinematics, it usually means your model or your signs need rechecking.
How we verify the algebra
Every run of the widget recomputes both roots, substitutes them back into the original equation and reports the residual — if |ax² + bx + c| > 10⁻⁹ we flag a warning. We reference AQA GCSE, Edexcel A-level textbooks and the CRC Standard Mathematical Tables. See our editorial policy and corrections policy — all calculations run in your browser.