The Cardano–Tartaglia formula and complex numbers.

Scipione dal Ferro (1465-1526) held the Chair of Arithmetic and Geometry at the University of Bologna and certainly must have met Pacioli who lectured at Bologna in 1501-2. dal Ferro is credited with solving cubic equations algebraically but the picture is somewhat more complicated. The problem was to find the roots by adding, subtracting, multiplying, dividing and taking roots of expressions in the coefficients. We believe that dal Ferro could only solve cubic equation of the form x³ + mx = n. In fact this is all that is required.

For, given the general cubic y³ - by² + cy - d = 0, put y = x + b/3 to get

x³ + mx = n where m = c - b²/3, n = d - bc/3 + 2b³/27.

However, without the Hindu's knowledge of negative numbers, dal Ferro would not have been able to use his solution of the one case to solve all cubic equations. Remarkably, dal Ferro solved this cubic equation around 1515 but kept his work a complete secret until just before his death, in 1526, when he revealed his method to his student Antonio Fior.

Fior was a mediocre mathematician and far less good at keeping secrets than dal Ferro. Soon rumours started to circulate in Bologna that the cubic equation had been solved. Nicolo of Brescia, known as Tartaglia meaning 'the stammerer', prompted by the rumours managed to solve equations of the form x³ + mx² = n and made no secret of his discovery.

Fior challenged Tartaglia to a public contest: the rules being that each gave the other 30 problems with 40 or 50 days in which to solve them, the winner being the one to solve most but a small prize was also offered for each problem. Tartaglia solved all Fior's problems in the space of 2 hours, for all the problems Fior had set were of the form x³ + mx = n as he believed Tartaglia would be unable to solve this type. However only 8 days before the problems were to be collected, Tartaglia had found the general method for all types of cubics.

News of Tartaglia's victory reached Girolamo Cardan in Milan where he was preparing to publish Practica Arithmeticae (1539). Cardan invited Tartaglia to visit him and, after much persuasion, made him divulge the secret of his solution of the cubic equation. This Tartaglia did, having made Cardan promise to keep it secret until Tartaglia had published it himself. Cardan did not keep his promise. In 1545 he published Ars Magna the first Latin treatise on algebra.

Here, in modern notation, is Cardan's solution of x³ + mx = n.

Notice that (a - b)³ + 3ab(a - b) = a³ - b³
so if a and b satisfy 3ab = m and a³ - b³ = n then a - b is a solution of x³ + mx = n.
But now b = m/3a so a³ - m³/27a³ = n,
i.e. a⁶ - na³ - m³/27 = 0.
This is a quadratic equation in a³, so solve for a³ using the usual formula for a quadratic.
Now a is found by taking cube roots and b can be found in a similar way (or using b=m/3a).
Then x = a - b is the solution to the cubic.

Cardan noticed something strange when he applied his formula to certain cubics. When solving x³ = 15x + 4 he obtained an expression involving -121. Cardan knew that you could not take the square root of a negative number yet he also knew that x = 4 was a solution to the equation. He wrote to Tartaglia on 4 August 1539 in an attempt to clear up the difficulty. Tartaglia certainly did not understand. In Ars Magna Cardan gives a calculation with 'complex numbers' to solve a similar problem but he really did not understand his own calculation which he says is as subtle as it is useless.