Cubic Equation Solver (Cardano's Formula)

Cubic roots by Cardano's method

Any cubic equation ax³ + bx² + cx + d = 0 reduces to its depressed form t³ + pt + q = 0 once you apply the substitution x = t - b/(3a), which clears the quadratic term. From there, Cardano's formula (Ars Magna, 1545) hands you one real root: t = ∛(-q/2 + √(q²/4 + p³/27)) + ∛(-q/2 - √(q²/4 + p³/27)).

Example: take t³ - 6t + 4 = 0, where p = -6 and q = 4. Then q²/4 + p³/27 = 4 - 8 = -4 < 0. That negative value flags the casus irreducibilis, the awkward case where three perfectly real roots come out of the cube roots of complex numbers. It was Bombelli (1572) who first worked formally with imaginary numbers to push through it.

Applications

You run into cubics more often than you might expect. They show up in chemical kinetics (equilibrium concentrations in third-order reactions), in solid mechanics (eigenvalues of 3×3 stress tensors and characteristic polynomials), in signal processing (cubic IIR filters), in the state equations of thermodynamics (van der Waals), and in computer graphics (Bézier cubics).

FAQ

What is the depressed form? Strip out the quadratic term with x = t - b/(3a) and you're left with t³ + pt + q = 0. That's the shape Cardano's formula tackles head-on.

What is the casus irreducibilis? It's what happens when q²/4 + p³/27 < 0. The cubic still has three distinct real roots, yet the radical formula detours through complex numbers to find them, so most people reach for the trigonometric form instead.

How are the other two roots found? Once you have one root r, divide the polynomial by (x - r). What's left is a quadratic, and solving that gives you the remaining two.

Why is the formula attributed to Cardano? Because he put it in print, in Ars Magna (1545). The history is messier than the name suggests: Tartaglia had shared the method with him in confidence, and del Ferro had cracked a partial case even earlier.