Solve any quadratic equation ax²+bx+c=0 instantly. Shows discriminant, real and complex roots, vertex coordinates, axis of symmetry, and parabola graph. Free for US Algebra 2 and UK GCSE/A-level students.
The quadratic formula is one of the most important and widely used formulas in all of mathematics. Whether you are a US student tackling Algebra 2 or a UK student preparing for GCSE or A-level maths, understanding how to solve quadratic equations is an essential skill. This free quadratic formula calculator solves any equation of the form ax² + bx + c = 0 instantly, showing all steps including the discriminant, roots, vertex, and parabola graph.
The quadratic formula is a universal method for solving any quadratic equation — that is, any polynomial equation of degree 2. The standard form of a quadratic equation is:
ax² + bx + c = 0
where a, b, and c are real numbers and a ≠ 0. The quadratic formula gives the solution (or solutions) directly:
x = (−b ± √(b² − 4ac)) / 2a
The ± symbol means the formula produces two solutions — one using + and one using −. These solutions are called the roots or zeros of the equation, and they represent the x-values where the parabola crosses the x-axis.
The quadratic formula is not just a formula to memorise — it can be derived from scratch using a technique called completing the square. Starting with ax² + bx + c = 0:
This derivation shows why the formula works and is often required knowledge in UK A-level and US Precalculus courses.
The discriminant is the expression inside the square root: Δ = b² − 4ac. The value of the discriminant tells you exactly what kind of roots the equation has before you even complete the calculation:
| Discriminant Value | Root Type | Graph Behaviour |
|---|---|---|
| Δ > 0 | Two distinct real roots | Parabola crosses x-axis at two points |
| Δ = 0 | One repeated real root | Parabola touches x-axis at exactly one point (vertex) |
| Δ < 0 | Two complex (imaginary) roots | Parabola does not touch the x-axis |
When Δ < 0, the roots involve imaginary numbers. The two complex roots are conjugates of each other: x = (−b ± i√|Δ|) / 2a, where i = √(−1).
Every parabola has a vertex — the highest or lowest point of the curve. The vertex coordinates (h, k) are given by:
The axis of symmetry is the vertical line x = h, which passes through the vertex and divides the parabola into two mirror-image halves.
Any quadratic ax² + bx + c can be rewritten in vertex form: y = a(x − h)² + k. This form makes it easy to identify the vertex (h, k) directly and to understand how the parabola is shifted and scaled relative to the basic y = x² curve.
Solve x² − 5x + 6 = 0. Here a = 1, b = −5, c = 6.
Solve x² − 6x + 9 = 0. Here a = 1, b = −6, c = 9.
Solve x² + 2x + 5 = 0. Here a = 1, b = 2, c = 5.
The sign of coefficient a determines whether the parabola opens upward or downward:
When an object is thrown, its height h at time t follows a quadratic equation: h = −½gt² + v₀t + h₀, where g is gravitational acceleration (9.8 m/s² in SI units or 32 ft/s² in imperial), v₀ is initial velocity, and h₀ is initial height. Setting h = 0 and solving gives the time when the object lands.
Quadratic functions model profit and revenue in many business scenarios. If profit = −2x² + 120x − 400 (where x is units sold), the maximum profit occurs at the vertex: x = −120/(2 × −2) = 30 units. Businesses use this to find the optimal production quantity.
Parabolic arches, satellite dishes, and bridge cables all follow parabolic curves. Engineers use quadratic equations to calculate load distributions, cable tensions, and optimal structural dimensions. The Gateway Arch in St. Louis follows a catenary curve that closely approximates a parabola.
When revenue R(x) = px and cost C(x) = ax² + bx + c, setting R = C gives a quadratic equation. Solving it finds the break-even quantities — how many units must be sold to cover costs.
In the UK, quadratic equations first appear in GCSE Mathematics at Higher Tier (grades 4–9). Students are expected to solve quadratics by factorisation, completing the square, and using the quadratic formula. The quadratic formula is provided on the GCSE formula sheet, so students must know how to apply it correctly.
At A-level, quadratics appear in Core Pure Mathematics (C1/C2 in legacy specs, now Pure 1/Pure 2). Students encounter the discriminant in detail, solve quadratic inequalities, and work with quadratic equations having complex roots in A-level Further Mathematics. The discriminant condition is particularly important for determining whether lines are tangent to curves.
In the United States, quadratic equations are introduced in Algebra 1 and studied in depth in Algebra 2. The quadratic formula is a core topic in the Common Core State Standards for Mathematics (CCSS-M), specifically under the Algebra domain: Reasoning with Equations and Inequalities. Students use the quadratic formula to solve equations, interpret the discriminant, and understand complex number solutions.
In Precalculus and AP Calculus courses, quadratic functions reappear in the study of polynomial functions, conic sections (parabolas), optimisation problems, and as building blocks for more complex functions. The SAT and ACT both regularly test quadratic equation solving.
Many quadratics can be factored quickly by inspection. For example, x² − 5x + 6 = (x − 2)(x − 3) = 0 gives roots x = 2 and x = 3 immediately. However, factoring only works when the roots are rational numbers. The quadratic formula works for any quadratic equation — including those with irrational or complex roots — making it the most general and reliable method.
The quadratic formula is x = (−b ± √(b² − 4ac)) / 2a. It solves any quadratic equation ax² + bx + c = 0, giving the values of x (called roots or zeros) where the parabola crosses the x-axis. The formula works for all quadratic equations, including those with irrational or complex roots.
The discriminant Δ = b² − 4ac tells you the nature of the roots. If Δ > 0, there are two distinct real roots. If Δ = 0, there is exactly one repeated real root (the parabola just touches the x-axis). If Δ < 0, there are two complex (imaginary) roots and the parabola does not cross the x-axis.
The vertex (h, k) is found using h = −b/2a for the x-coordinate and k = c − b²/4a (or substitute h back into the equation) for the y-coordinate. The vertex is the minimum point if a > 0 and the maximum point if a < 0.
Completing the square is a method of rewriting a quadratic expression as a perfect square plus a constant: ax² + bx + c = a(x + b/2a)² + (c − b²/4a). It is used to derive the quadratic formula, find the vertex form of a parabola, and solve quadratic equations without the formula. It is a key technique in UK GCSE and A-level maths.
Yes. When the discriminant b² − 4ac is negative, the square root produces an imaginary number. The two complex roots are conjugates: x = (−b + i√|Δ|) / 2a and x = (−b − i√|Δ|) / 2a, where i = √(−1). Complex roots appear in pairs and are studied in UK A-level Further Maths and US Algebra 2/Precalculus.
If the roots of ax² + bx + c = 0 are x₁ and x₂, then the quadratic factors as a(x − x₁)(x − x₂). Roots are the x-values that make the expression equal to zero; factors are the corresponding linear expressions. Finding the roots is equivalent to factoring — the two methods give the same information in different forms.
In projectile motion, height is modelled as h(t) = −½gt² + v₀t + h₀. Setting h = 0 and applying the quadratic formula gives the time(s) when the object is at ground level. The positive root gives the landing time; the negative root is rejected as physically meaningless. This application appears in both GCSE Physics and US Physics courses.
Yes. The quadratic formula is provided on the GCSE Mathematics Higher Tier formula sheet in England, Wales, and Northern Ireland. Students must be able to apply it to solve quadratic equations that cannot be easily factored. Completing the square and using the formula to solve problems involving the discriminant are both assessed in GCSE and A-level examinations.
Results are calculated using standard mathematical formulas for educational purposes only. Always verify solutions by substituting back into the original equation. Complex root notation may differ between educational systems.
A quadratic equation has the form ax² + bx + c = 0. The quadratic formula solves it: x = (−b ± √(b² − 4ac)) / 2a. The expression under the square root, b² − 4ac, is called the discriminant. If positive → two real solutions; if zero → one repeated solution; if negative → two complex solutions. Our calculator shows each step including discriminant calculation.
a = 1, b = −5, c = 6. Discriminant = (−5)² − 4(1)(6) = 25 − 24 = 1 (positive → two real solutions). x = (5 ± √1) / 2 = (5 ± 1) / 2. So x = 3 or x = 2. Verify: 3² − 5(3) + 6 = 9 − 15 + 6 = 0 ✓; 2² − 5(2) + 6 = 4 − 10 + 6 = 0 ✓.
It is derived from completing the square on the general form. Starting from ax² + bx + c = 0: divide by a → x² + (b/a)x + c/a = 0. Complete the square: (x + b/2a)² = (b² − 4ac)/(4a²). Take square root and isolate x. The same algebraic manipulation, written more compactly, gives the formula every secondary-school student learns.
Factoring works for quadratics with integer roots — fast when applicable (x² − 5x + 6 → (x−2)(x−3)). Completing the square works always but takes more steps. Quadratic formula works always and is mechanical — best for non-integer or messy coefficients. Most A-Level / Algebra 2 questions test all three methods.
A quadratic in standard form ax² + bx + c can be rewritten in vertex form a(x − h)² + k where h = −b/(2a) and k = c − b²/(4a). The vertex (h, k) is the parabola's minimum (if a > 0) or maximum (if a < 0). Critical for optimisation problems and revenue/cost modelling.
x = (−b ± √(b² − 4ac)) / 2a. Solves ax² + bx + c = 0 for any a, b, c.
The expression b² − 4ac inside the square root. Positive → 2 real roots, zero → 1 repeated root, negative → 2 complex roots.
Try factoring first if coefficients are small integers. If factoring isn't obvious within 30 seconds, switch to the formula.
Yes — if the discriminant is negative. The √ of a negative gives an imaginary result, so the roots are complex conjugates.
a(x − h)² + k, where (h, k) is the parabola's vertex. h = −b/(2a); k = c − b²/(4a).