Quadratic Equation Solver
Any rearranged quadratic solved exactly three ways, with discriminant, vertex and graph.
Solution
This is the last answer worked out. Fix the input above to update it.
Read as
Standard form
Roots
Discriminant and nature of the roots
Step-by-step
Vertex form and axis of symmetry
Factored form
Sum and product of the roots
About the Quadratic Equation Solver
Type a quadratic equation the way it appears in your book — 2x² = 3x + 5, (x − 1)(x + 2) = 4, x(x + 3) = 10 or even x + 1/x = 5/2 — or enter the coefficients a, b and c. The solver rearranges it into the standard form ax² + bx + c = 0, works out the discriminant and the nature of the roots, and gives the roots exactly: as fractions, simplified surds such as (−1 + √33)/4, or complex numbers such as −1 ± 2i, each with its decimal value.
The working is shown three ways — the quadratic formula, factorisation by splitting the middle term, and completing the square — together with the sum and product of the roots, the vertex form, the axis of symmetry and a graph of the parabola with its vertex and x-intercepts marked. All arithmetic is exact (decimals such as 0.5 are read as 1/2), and nothing you type leaves your browser.
How to use it
- Type the equation in any form, using ^ or ² for powers (x^2 or x²). Or choose “Enter a, b, c” and type the three coefficients — whole numbers, decimals or fractions such as −3/4.
- Press Enter or Solve (the answer also updates as you type). Check the “Read as” and “Standard form” lines to see how your equation was understood.
- Read the roots and the nature of the roots from the discriminant, then pick a method — quadratic formula, factorisation or completing the square — to see every step.
- Use the vertex form, axis of symmetry, sum and product of the roots and the graph for the rest of the question. Copy or download everything as text.
Examples
2x² − 3x − 5 = 0
x = −1 or x = 5/2
D = (−3)² − 4·2·(−5) = 49 = 7², a perfect square, so it factorises: (x + 1)(2x − 5) = 0.
x² − 2x − 4 = 0
x = 1 ± √5 ≈ −1.236 and 3.236
D = 20 is not a perfect square; √20 = 2√5, so x = (2 ± 2√5)/2 = 1 ± √5.
4x² − 12x + 9 = 0
x = 3/2 (repeated)
D = 144 − 144 = 0; the left side is the perfect square (2x − 3)².
x² + 2x + 5 = 0
x = −1 ± 2i
D = 4 − 20 = −16 < 0: the parabola stays above the x-axis; its vertex is (−1, 4).
x + 1/x = 5/2
x = 1/2 or x = 2
Multiplying by x (x ≠ 0) and rearranging gives x² − (5/2)x + 1 = 0.
x² − 4x + 3 = 0
y = (x − 2)² − 1, vertex (2, −1), axis x = 2
h = −b/(2a) = 2 and k = −D/(4a) = −4/4 = −1.
Common uses
- Checking homework and exam answers for Class 10 quadratic equations, with the method your teacher asked for.
- Finding whether a quadratic has real roots (and how many) from the discriminant before solving it.
- Getting the vertex, axis of symmetry and intercepts of a parabola for a graphing question.
- Writing the equation whose roots have a given sum and product, or checking α + β and αβ.
The formulas used
- Standard form: ax² + bx + c = 0 with a ≠ 0.
- Quadratic formula: x = (−b ± √(b² − 4ac)) / (2a).
- Discriminant: D = b² − 4ac. If D > 0 the equation has two distinct real roots, if D = 0 two equal real roots, and if D < 0 no real roots (NCERT Class 10 Mathematics, Chapter 4, Quadratic Equations). With rational coefficients the roots are rational exactly when D is a perfect square.
- Sum and product of the roots: α + β = −b/a and αβ = c/a (the relationship between the zeroes and coefficients of a quadratic polynomial, NCERT Class 10, Chapter 2).
- Vertex form: y = a(x − h)² + k with h = −b/(2a) and k = c − b²/(4a) = −D/(4a). The vertex is (h, k) and the axis of symmetry is the line x = h.
The three methods
Quadratic formula — always works. Substitute a, b and c, simplify the square root (√20 = 2√5, √−16 = 4i) and cancel common factors.
Factorisation by splitting the middle term — find two numbers whose product is ac and whose sum is b, split bx into two terms, group in pairs and take out the common bracket. Such whole numbers exist only when the discriminant is a perfect square, so the tool says when a quadratic cannot be factorised with rational numbers. Fractions are cleared first (multiplying by the lowest common denominator) so the numbers are whole.
Completing the square — divide by a, move the constant to the right, add the square of half the coefficient of x to both sides, write the left side as (x + b/2a)², and take square roots of both sides.
Typing equations
- Powers:
x^2orx²; multiplication can be left out:2x,3(x + 1),x(x − 4). - Fractions and decimals:
x/2,(2/3)x,0.25x^2— decimals are converted to exact fractions. - Any letter can be the unknown:
t^2 − 5t + 6 = 0. - Without “=”, the expression is set equal to 0. An unknown in a denominator is allowed when clearing it leaves a quadratic; values that make a denominator zero are rejected.
Limitations
- Coefficients must be rational numbers (whole numbers, decimals or fractions). Coefficients such as √2 or π are not accepted.
- One unknown only, and the equation must be of degree 2 after rearranging (degree 1 is solved as a linear equation). For cubic and higher-degree equations use the Equation Solver.
- Decimal values are rounded to 10 significant digits; the exact forms (fractions, surds and complex numbers) are exact.
Privacy
Everything happens in your browser. What you enter or open here is not uploaded or stored by MySmartCoPilot.
Frequently asked questions
What does the discriminant tell me?
D = b² − 4ac decides the nature of the roots without solving. D > 0: two different real roots (rational if D is a perfect square, otherwise a pair of conjugate surds). D = 0: one repeated real root, x = −b/(2a). D < 0: no real roots — the parabola does not touch the x-axis, and the two roots are complex conjugates.
How do I know if a quadratic can be factorised?
With whole-number (or rational) coefficients, ax² + bx + c factorises into rational linear factors exactly when the discriminant is a perfect square. For 2x² − 3x − 5, D = 49 = 7², so it factorises as (x + 1)(2x − 5); for x² − 2x − 4, D = 20 is not a perfect square, so use the formula.
Why are the roots of x² + 2x + 5 = 0 complex?
Its discriminant is 4 − 20 = −16, which is negative, so there is no real number whose square gives the required value. The formula gives x = (−2 ± √−16)/2 = (−2 ± 4i)/2 = −1 ± 2i, where i = √−1. On the graph, the parabola has its lowest point at (−1, 4), above the x-axis.
How do I find the sum and product of the roots without solving?
For ax² + bx + c = 0 the roots α and β satisfy α + β = −b/a and αβ = c/a. For 2x² − 3x − 5 = 0, the sum is 3/2 and the product is −5/2 — which matches −1 + 5/2 and (−1)(5/2).
What is the vertex form used for?
y = a(x − h)² + k shows the vertex (h, k) directly: the lowest point of the parabola when a > 0 or the highest when a < 0, so k is the minimum or maximum value. The axis of symmetry is the vertical line x = h, halfway between the two roots.
What happens if a = 0?
Then there is no x² term and the equation is linear, bx + c = 0, with the single solution x = −c/b. The tool says so and solves it; if b is 0 too, the equation is either always true or never true.
Can I type the equation in any order?
Yes. Terms can be on both sides, in brackets or multiplied out: 2x² = 3x + 5, (x − 1)(x + 2) = 4 and x(x + 3) = 10 all work. The tool expands and collects everything into ax² + bx + c = 0 and shows the result so you can check it.