Two rational roots
For x²−5x+6=0, Δ=(−5)²−4(1)(6)=1. Since 1 is positive and a perfect square, the roots are distinct, real, and rational.
Instantly see whether the quadratic has two distinct real roots, one repeated real root, or two complex-conjugate roots—and whether real roots are rational.
Because Δ is positive and its square root is rational.
For ax² + bx + c = 0, with a ≠ 0, the discriminant is the expression beneath the square-root sign in the quadratic formula.
Δ = b² − 4ac and x = (−b ± √Δ)/(2a)
| Value of Δ | Roots over the real numbers | Graph of y = ax² + bx + c |
|---|---|---|
| Δ > 0 | Two distinct real roots | Crosses the x-axis twice |
| Δ = 0 | One repeated real root | Touches the x-axis once at the vertex |
| Δ < 0 | No real roots; two complex roots | Does not meet the x-axis |
When a, b, and c are rational, a nonnegative discriminant has a rational square root only when it is a perfect square of a rational number. Then the roots are rational. If Δ > 0 but √Δ is irrational, both real roots are irrational.
For x²−5x+6=0, Δ=(−5)²−4(1)(6)=1. Since 1 is positive and a perfect square, the roots are distinct, real, and rational.
For x²+2x−1=0, Δ=2²−4(1)(−1)=8. It is positive but not a perfect square, so the two real roots are irrational.
For x²−6x+9=0, Δ=36−36=0. The parabola touches the x-axis at the repeated root x=3.
For x²+4x+8=0, Δ=16−32=−16. There are no real x-intercepts and the roots are −2±2i.
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For ax²+bx+c=0, the discriminant is Δ=b²−4ac. It is the quantity under the square root in the quadratic formula.
A positive discriminant gives two distinct real roots, zero gives one repeated real root, and a negative discriminant gives two non-real complex-conjugate roots.
For rational coefficients, real roots are rational exactly when the discriminant is a square of a rational number. A positive nonsquare discriminant produces two irrational real roots.
Yes. Fractional or decimal coefficients can produce a fractional discriminant. Its sign still classifies the roots normally.
The expression is no longer quadratic, so this classification does not apply. Use the Linear Equation Solver when a is zero and b is not.
There are no real roots, but there are two roots over the complex numbers. They form a complex-conjugate pair.