Quadratic Equation Calculator

How to use this calculator

1. Enter coefficients with their signsFor x² − 5x + 6 enter a = 1, b = −5, c = 6. A missing term means that coefficient is zero — x² − 9 has b = 0.
2. Rearrange to equal zero firstThe formula assumes the equation is set to zero. x² = 5x − 6 must become x² − 5x + 6 = 0 before the coefficients can be read off.
3. Read the discriminant before the rootsIt tells you what kind of answer to expect. A positive perfect square means the quadratic factorises neatly; a positive non-square means irrational roots.
4. Use the Vieta checkThe last two rows give the sum and product of the roots. Verifying against them catches most sign errors in a hand calculation.

The discriminant decides everything

The expression under the square root sign is the discriminant, and checking it first tells you the shape of the answer before you do any work.

If it is positive, the square root is a real number, so the ± gives two different values and the quadratic has two distinct real roots. Graphically, the parabola crosses the x-axis twice.

If it is exactly zero, the ± adds and subtracts nothing, so both roots collapse to the same value. The parabola touches the x-axis at a single point, which is its vertex.

If it is negative, the square root of a negative number is not real, so there are no real roots. The parabola sits entirely above or entirely below the axis. The roots still exist as a complex conjugate pair, which this calculator reports rather than declining to answer.

There is a further distinction worth noticing when the discriminant is positive. If it is a perfect square, the square root is a whole number and the roots are rational — which means the quadratic factorises neatly with integer coefficients. If it is positive but not a perfect square, the roots are irrational and factorising by inspection will not work.

That is a practical time-saver in an exam. Computing b² − 4ac takes a few seconds and tells you immediately whether to spend time looking for factors or go straight to the formula. Students who try to factorise x² − 5x + 3 by inspection are looking for something that does not exist, and the discriminant of 13 would have told them so.

Worked example: x² − 5x + 6 = 0

Here a = 1, b = −5 and c = 6.

Discriminant: (−5)² − 4(1)(6) = 25 − 24 = 1.

Positive, so two distinct real roots. And it is a perfect square, so the roots will be rational and the quadratic factorises.

Apply the formula: x = (5 ± √1) ÷ 2 = (5 ± 1) ÷ 2.

That gives x = 6 ÷ 2 = 3 and x = 4 ÷ 2 = 2.

Factored form: (x − 3)(x − 2). Expanding it back gives x² − 5x + 6, which confirms the roots.

The vertex sits at x = −b ÷ 2a = 5 ÷ 2 = 2.5, exactly midway between 2 and 3 as it must be. Substituting back: y = 2.5² − 5(2.5) + 6 = 6.25 − 12.5 + 6 = −0.25. So the vertex is (2.5, −0.25), and since a is positive the parabola opens upward and that is a minimum.

Now the Vieta check. The roots should sum to −b/a = 5, and 3 + 2 = 5 ✓. They should multiply to c/a = 6, and 3 × 2 = 6 ✓.

That check takes seconds and catches the most common error in this topic, which is dropping a minus sign somewhere in −b. If your roots do not satisfy both relationships, go back and look at the signs.

What the discriminant tells you

The first row is what exam questions are usually built around, because the answers come out cleanly. Spotting a perfect-square discriminant early tells you factorising is worth attempting; anything else and the formula is the faster route. A discriminant of zero is also worth recognising, because it means the quadratic is a perfect square such as x² − 6x + 9 = (x − 3)².

Completing the square, and where quadratics turn up

The quadratic formula is not magic — it is the result of completing the square on the general form once, so the derivation does not have to be repeated for every equation.

Completing the square is still worth knowing in its own right, because it produces the vertex form directly. Rewriting ax² + bx + c as a(x − h)² + k puts the vertex at (h, k), which the standard form hides. That is the fastest route to a maximum or minimum without calculus.

The vertex is often what a problem actually wants. Maximum profit, minimum cost, the peak of a projectile, the shortest distance — all of these are vertex questions dressed up as word problems, and the answer is x = −b ÷ 2a rather than either root.

Projectile motion is the classic physical case. Vertical height under constant gravity is quadratic in time, so the roots are the two moments the object is at ground level and the vertex is the peak. A negative root usually means a time before the throw, which is mathematically valid and physically meaningless — worth checking rather than reporting.

Quadratics also appear in area problems where a dimension is unknown, in optimisation of a fence or enclosure, in the intersection of a line with a circle, and throughout kinematics.

One recurring trap in word problems: both roots are mathematically correct, but only one usually makes physical sense. A negative length, a negative time, or a number of items that is not a whole number should be discarded with a sentence explaining why. Examiners look for that reasoning, and simply presenting both roots loses the mark.

For higher powers, there is a cubic formula and a quartic formula, both far more involved. Above degree four, the Abel-Ruffini theorem proves no general formula in radicals exists at all — which is why numerical methods take over.

Assumptions and limitations

• Real coefficients only — complex inputs are not accepted
• Roots are computed numerically, so irrational roots show as decimals rather than surds
• A zero leading coefficient is handled as a linear equation, not a quadratic
• Does not factorise irrational roots into exact surd form
• Does not solve simultaneous or higher-degree equations
• Very large or very small coefficients can lose precision in floating-point arithmetic
• Does not judge which root is physically meaningful in a word problem

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