About this tool

Quadratic Equation Solver

Solve ax² + bx + c = 0 — real or complex roots, discriminant and vertex.

This Quadratic Equation Solver takes the coefficients a, b and c of ax² + bx + c = 0 and returns the roots from the quadratic formula x = (-b ± √(b² - 4ac)) / 2a, along with the discriminant, the nature of the roots, and the vertex at (-b/2a, f(-b/2a)). Students and anyone checking algebra homework get real roots to three decimals when the discriminant is positive, a double root when it is exactly zero, and a ± bi complex pair when it is negative. Enter a = 0 and it tells you the equation is not quadratic rather than dividing by zero.

How to use Quadratic Equation Solver

1. 1

   Open Quadratic Equation Solver on AltFTool — it loads instantly in your browser.

2. 2

   Enter the coefficients into the three input fields labelled a, b and c to define your quadratic equation; they start at 1, -3 and 2, and setting a to 0 replaces the result with the message a can't be zero (not quadratic)

3. 3

   There is no Solve button — the Result panel recalculates as you type, printing x₁ and x₂ for a positive discriminant, a single x marked (double root) when it is zero, or the x = h ± ki form when it is negative

4. 4

   Review the cards under the result — Discriminant, Nature (two real roots, one real root or two complex roots) and Vertex as an (x, y) pair — then use Copy or Download to save the whole solution as quadratic-equation-solver.txt

Why use Quadratic Equation Solver

• Complex roots are solved, not refused

  When b² - 4ac is negative the answer is returned as -b/2a ± (√|D|)/2a i, instead of stopping at "no real solution".

• Discriminant and vertex come with the roots

  Each run reports b² - 4ac, whether that means two real, one real or two complex roots, and the turning point of the parabola.

• Degenerate input is caught

  If a is zero the equation is linear, and the solver says so instead of returning infinity or NaN from the 2a denominator.

Common use cases

• You have worked x² - 3x + 2 = 0 by factoring and want to confirm the roots really are 1 and 2 before handing in the sheet.
• A projectile problem gives you a height equation and you need the vertex to find the peak, so you read the (-b/2a, f) pair straight off the result instead of completing the square.
• Your discriminant came out negative and you want the complex roots written properly as a real part plus an imaginary part rather than being told "no solution".

Frequently asked questions