What formula does this calculator use?

It uses two complementary methods. When the discriminant Δ = −4p³ − 27q² is positive (three real roots), it applies the trigonometric de Moivre method, which avoids complex arithmetic entirely. When Δ is negative (one real root plus a complex pair), it uses Cardano's formula with cube roots. When Δ equals zero, the repeated-root formula applies directly.

What is the depressed cubic?

Any cubic ax³ + bx² + cx + d = 0 can be transformed into the simpler form t³ + pt + q = 0 by substituting x = t − b/(3a). This shift removes the x² term. The coefficients p and q depend only on a, b, c, d and are shown in the results panel. Solving the depressed form first, then shifting back, is the foundation of every classical solution method.

What does the discriminant tell me?

The discriminant of a depressed cubic is Δ = −4p³ − 27q². If Δ is greater than 0 the polynomial has three distinct real roots; if Δ equals 0 it has a repeated real root (a double root and a simple root, or a triple root if p and q are both zero); if Δ is less than 0 it has one real root and two non-real complex conjugate roots.

Can a cubic equation have no real roots?

No. Every cubic polynomial with real coefficients has at least one real root, because a continuous function of odd degree must cross the x-axis. When Δ is less than 0 there is exactly one real root; the other two are a complex conjugate pair.

How accurate are the results?

The solver uses JavaScript double-precision floating-point (64-bit IEEE 754), giving roughly 15–16 significant digits. For typical classroom coefficients (integers up to a few hundred) the answers are accurate to at least 10 significant figures. Large or near-singular inputs (for example when two roots are almost identical) may show slightly reduced precision because of cancellation, which is a fundamental limitation of floating-point arithmetic rather than a flaw in the algorithm.

What happens if I set a to zero?

The tool detects this and automatically falls back to the appropriate lower-degree solver. If a = 0 but b is non-zero it solves the quadratic bx² + cx + d = 0 and reports the discriminant. If a = 0 and b = 0 it solves the linear equation cx + d = 0. If all three of a, b and c are zero it reports whether the equation is always true or always false.

Cubic Equation Solver

A cubic equation has the general form ax³ + bx² + cx + d = 0, where a is non-zero. Unlike quadratics, a cubic always has at least one real root — its graph must cross the x-axis at least once. The tool above solves any such equation immediately in your browser, showing not only the roots but every algebraic step that leads to them.

How it works

Step 1 — Depress the cubic

The first move is to eliminate the x² term using the substitution x = t − b / (3a). After this shift the equation becomes the depressed cubic t³ + pt + q = 0, where

p = (3ac − b²) / (3a²)

q = (2b³ − 9abc + 27a²d) / (27a³)

Working in the depressed form is far cleaner because it has only two free parameters instead of four.

Step 2 — Evaluate the discriminant

The discriminant of the depressed cubic is

Δ = −4p³ − 27q²

This single number classifies the roots completely:

Δ	Root structure
Δ > 0	Three distinct real roots
Δ = 0	A double root and a simple root (or a triple root when p = q = 0)
Δ (negative)	One real root plus two complex conjugates

Step 3 — Solve by the right method

Three real roots (Δ > 0): The tool uses the trigonometric de Moivre method. Setting m = 2·sqrt(−p / 3) and computing θ = (1/3)·arccos(3q / (pm)), the three roots are t_k = m·cos(θ − 2kπ/3) for k = 0, 1, 2. This avoids complex arithmetic even though the intermediate expressions might look imaginary with other approaches — an issue historically known as casus irreducibilis.

One real root (Δ negative): Cardano’s formula gives the real root as u + v, where u = cbrt(−q/2 + sqrt(D)) and v = cbrt(−q/2 − sqrt(D)) with D = (q/2)² + (p/3)³ > 0. The complex conjugate pair follows from Vieta’s formulas.

Repeated root (Δ = 0): The double root of the depressed cubic is t = ∛(q/2) and the simple root is t = −2·∛(q/2), giving x = t − shift for each.

Step 4 — Shift back

Every root t of the depressed cubic maps back to a root of the original via x = t − b / (3a).

Worked example

Solve x³ − 6x² + 11x − 6 = 0.

Here a = 1, b = −6, c = 11, d = −6. The depressed coefficients are p = −1, q = 0. Because p is negative and q is zero, the discriminant Δ = −4(−1)³ − 27(0)² = 4 > 0, so there are three distinct real roots. The trigonometric method gives t₁ = 2, t₂ = −1, t₃ = −1 (roughly, before shifting). Shifting back by −b/(3a) = 2 yields x₁ = 1, x₂ = 2, x₃ = 3 — which you can verify directly by substitution: 1 − 6 + 11 − 6 = 0.

For comparison, x³ + x + 2 = 0 (p = 1, q = 2, Δ = −4 − 108 = −112, negative) has one real root x = −1 and two complex conjugate roots.

Formula note

Cardano published his formula in Ars Magna in 1545, making it one of the first explicit solutions for a polynomial of degree higher than two. The trigonometric form used here for three real roots was refined over the following century. The discriminant Δ = −4p³ − 27q² is the standard expression for a monic depressed cubic; it factors as the square of the Vandermonde product (r₁ − r₂)(r₁ − r₃)(r₂ − r₃) times a² (the squared leading coefficient), which is why Δ = 0 forces at least two roots to coincide.