Quadratic Equation Solver

Two of the three answers a quadratic can give are the ones people forget to look for. A discriminant of exactly 0 folds both roots into a single value counted twice. A negative discriminant pushes them off the real line altogether, and a solver that only prints decimals will quietly show you nothing at all. Quadratic Equation Solver names which of the three cases you are in before it prints a number.

Roots arrive in radical form first. Enter a = 1, b = -3, c = -5 and you get (3 + √29) / 2 rather than a truncated 4.1925824. One power-of-ten scaling clears any decimal points you typed, the three coefficients are then held as BigInt integers, and b² − 4ac is evaluated with no rounding anywhere in it. Square factors come out of the radicand before it is printed, so √51340 is shown as 2√12835.

Below the result cards sits a sketch of the parabola those coefficients describe, with the vertex marked in terracotta and every real root labelled where the curve meets the axis. All of it runs in the tab you have open. No coefficient you type is sent anywhere.

Walkthrough: A Ball Thrown at 22 m/s

Load Sample fills the three boxes with -4.9, 22 and 1.5, which is the height in metres of a ball leaving a hand 1.5 m off the ground at 22 m/s. Here is what the working panel shows, step by step.

• The discriminant comes out as 484 − (−29.4) = 513.4. It is positive, so the ball is at ground level at two separate instants.
• Multiplying every coefficient by −10 clears the decimal and makes a positive, giving 49x² − 220x − 15 = 0 with the same two roots and an integer discriminant of 51340.
• That integer factors as 2² × 12835, so the radical simplifies and both roots reduce to (110 ± √12835) / 49.
• The decimal hints under the cards read 4.5569726 and -0.06717671. Only the first is a real instant; the negative root is where the parabola would have crossed had the throw started from the ground.
• The Vertex card answers a question you did not ask: (110/49, 2567/98), or roughly 2.24 s to the top of the arc at a height of 26.19 m.

The Formula, and the Three Things It Hands You Free

Everything on the page comes from b² − 4ac and the two coefficients above it. Nothing is fitted, sampled or approximated.

• The roots are (−b ± √D) / 2a, reduced by the greatest common divisor of the constant, the radical coefficient and the denominator, which is what turns (220 ± 2√12835) / 98 into its lowest terms.
• The vertex falls out as h = −b / 2a and k = −D / 4a, both kept as exact fractions. No second calculation is run and no completing-the-square detour is needed.
• Sum and product of the roots are −b/a and c/a. Those two hold even in the complex case, which makes them the fastest way to check a result by hand.
• Factored form appears only when D is a perfect square, because that is exactly when the trinomial splits over the rationals. With 1, -6, 9 you get (x − 3)²; with 1, 0, 2 no factored card appears, and that absence is the answer.
• Leaving a box empty counts it as zero, so 3x² − 12 = 0 needs nothing typed into b at all. Most people fill it with a 0 by hand on the first visit.

Precision When b² Dwarfs 4ac

The radical forms are exact. The decimal hints beneath them are double precision, and there is one classic case where writing the formula out longhand destroys them.

• For x² + 100000000x + 1 = 0, the naive expression (−b + √D) / 2a subtracts two numbers that agree to 16 digits and returns -7.4505806e-9. The true root is -1e-8, so the first significant figure is already wrong. This solver computes the well-conditioned root first and recovers the other from the product rule, which returns -1e-8.
• Each coefficient is limited to 12 significant digits. A 13-digit entry turns the box red under Each coefficient is capped at 12 digits — this one has 13.
• Square factors are stripped from the radicand by trial division up to 20000, so any leftover below 400 million is provably square-free. Above that a note appears saying the surd may not be fully reduced, rather than the tool implying a simplification it did not verify.
• Typing a coefficient the parser cannot read stops the run with Fix the highlighted coefficient before the solver can run. rather than silently treating it as zero.
• With a set to 0 there is no parabola, so the tool solves the linear equation that remains and says so instead of dividing by zero.