Quadratic Equation Solver
Solve a quadratic typed as written, like x^2 + 3x = 10 or (x-2)(x+3)=0: exact and decimal roots, factored form, vertex and full working.
Type a quadratic equation the way it appears in your homework or notes, and this solver rearranges it, solves it and shows the working. It accepts terms on both sides of the equals sign, brackets such as (x − 2)(x + 3), implied multiplication like 3x, and powers written as x^2 or x². You get exact and decimal roots, the factored form and the vertex of the matching parabola.
How to use this tool
- Type the equation, for example
x^2 + 3x = 10or(2x + 1)^2 = 9. - The solver expands brackets and moves every term to the left, giving ax² + bx + c = 0.
- Read the exact solutions first, then the decimal roots, discriminant and factored form.
- Use the working table to follow the rearrangement and the quadratic formula line by line.
Formula
x = (−b ± √(b² − 4ac)) / 2a
After rearranging to ax² + bx + c = 0, a, b and c are the coefficients of x², x and the constant term. The discriminant b² − 4ac decides the kind of answer: positive gives two real roots, zero gives one repeated root, negative gives a pair of complex roots.
Example
For x² + 3x = 10, subtract 10 from both sides: x² + 3x − 10 = 0, so a = 1, b = 3 and c = −10. The discriminant is 3² − 4 × 1 × (−10) = 49 and √49 = 7, so x = (−3 ± 7) / 2. That gives x = −5 or x = 2, and the factored form is (x − 2)(x + 5). The parabola y = x² + 3x − 10 has its vertex at (−3/2, −49/4).
Frequently asked questions
What kinds of equation can I type?
Anything that expands to a polynomial of degree 2 or less in one variable: x^2 = 5x, 3(x − 1)^2 = 12, 0.5t² − t = 4 or x(x + 4) = 21. Dividing by a number is fine (x^2/2); dividing by an expression containing the variable is not.
What happens if the x² terms cancel?
Then it is a linear equation, and the solver reports its single solution instead: x² + 2x = x² + 8 gives x = 4. If everything cancels, it tells you whether every number works or none does.
Why are some roots shown with i?
A negative discriminant means the parabola never crosses the x-axis, so there is no real solution. The two roots are complex conjugates such as −1 ± 2i, where i = √−1.
Can it solve cubic equations?
No. If the expanded equation has an x³ or higher term, the solver stops and says so, because the quadratic formula applies only to degree 2.
Last updated: 2026-10-30