What this calculator does
A zero (or root) of a function f(x) is any value of x where f(x) = 0 — the point where the graph crosses the x-axis. This tool finds the zeros of a quadratic function written in standard form \(f(x) = ax^2 + bx + c\). Enter the coefficients a, b and c and it returns the real or complex roots along with the discriminant.
How to use it
Type the three coefficients. For example, for \(f(x) = x^2 - 3x + 2\), enter a = 1, b = −3, c = 2. If a = 0 the equation becomes linear (bx + c = 0) and the calculator returns its single root. The discriminant row tells you whether the roots are two distinct reals, one repeated real, or a complex conjugate pair.
The formula explained
The zeros are found with the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
The quantity under the square root, \(\Delta = b^2 - 4ac\), is the discriminant. When \(\Delta > 0\) there are two distinct real zeros; when \(\Delta = 0\) there is one repeated real zero; when \(\Delta < 0\) the roots are complex, written as \(a \pm bi\).
Worked example
Solve \(f(x) = x^2 - 3x + 2 = 0\). Here a = 1, b = −3, c = 2. The discriminant is \((-3)^2 - 4(1)(2) = 9 - 8 = 1\). So $$x = \frac{3 \pm \sqrt{1}}{2} = \frac{3 \pm 1}{2},$$ giving \(x_1 = 2\) and \(x_2 = 1\). Indeed \(f(x) = (x - 1)(x - 2)\).
FAQ
What does a negative discriminant mean? The parabola never touches the x-axis, so there are no real zeros — the roots are a complex conjugate pair.
Can I use it for a linear function? Yes. Set a = 0 and it solves bx + c = 0, returning \(x = -c/b\).
Why do I sometimes get one root instead of two? When the discriminant is exactly zero, both roots coincide, so the quadratic has a single (repeated) zero at \(x = -b/(2a)\).
