Math Tools
Solve linear and quadratic equations. Enter coefficients to find roots instantly.
Quadratic equation: ax² + bx + c = 0
Roots
x₁ = 2
x₂ = 1
Two distinct real roots.
Quadratic equations must have a non-zero coefficient a. Complex roots appear as a ± bi.
An equation solver is a mathematical tool that finds the value(s) of unknown variables (usually x or y) that make an equation true. For example, in the equation 2x + 5 = 15, the solver determines that x = 5 because 2(5) + 5 = 15. It works with linear equations (first-degree), quadratic equations (second-degree), and higher-order polynomials.
The solver rearranges the equation to isolate the variable step by step. For linear equations, it moves constants to one side and variables to the other, then divides by the coefficient. For quadratic equations (ax² + bx + c = 0), it applies the quadratic formula or factors the expression. The tool displays each step so you understand not just the answer, but the method used to find it.
Quadratic formula: x = (-b ± √(b² - 4ac)) / 2aFor any quadratic equation ax² + bx + c = 0, substitute the coefficients a, b, and c into this formula. The ± symbol means there are usually two solutions (roots) for a quadratic equation. The expression under the square root (b² - 4ac) is called the discriminant and tells you how many real solutions exist.
| Input | Result | Notes |
|---|---|---|
| Linear: 3x + 7 = 22 | x = 5 | Subtract 7 from both sides to get 3x = 15, then divide by 3. |
| Quadratic: x² - 5x + 6 = 0 | x = 2 or x = 3 | Factors as (x - 2)(x - 3) = 0, so both 2 and 3 satisfy the equation. |
| Two-variable: 2x + 3y = 12 with y = 2 | x = 3 | Substitute y = 2 to get 2x + 6 = 12, then solve for x. |
Yes. Enter them as they are (e.g., 0.5x + 2 = 7 or x/4 + 3 = 5). The solver will work with both.
Linear equations have one solution and involve variables with exponent 1. Quadratic equations involve variables squared (exponent 2) and typically have two solutions. The solver uses different methods for each.
This means the discriminant (b² - 4ac) is negative, so there are no real-number answers — only complex numbers. This happens when the parabola doesn't cross the x-axis.
Yes, if you specify which variable to solve for. For example, in 2x + 3y = 12, you can solve for x in terms of y, or vice versa.
Absolutely. The step-by-step breakdown helps students understand the process, not just memorize answers. Teachers often use such tools to check work and explain methods.
Very advanced equations like trigonometric, exponential, or transcendental equations may require specialist tools, though basic versions might work here.