# How Tartaglia Solved the Cubic Equation - The Other Cubic Equations

Author(s):
Friedrich Katscher

To solve the ten cubic equations with the quadratic term, this term has to be removed by transforming it into a linear term, and the transformed equation then can be solved by the formula of Scipione dal Ferro and Tartaglia, which today is unjustly called Cardano's formula because he was the first to publish it, in his Ars magna in 1545.

Tartaglia did not find a way to solve the ten cubic equations with the quadratic term. It is true that he printed the solution of the equation .1.cubo piu .6.censi equal à .100. (x3+6x2=100) in his book Quesiti et inventioni diverse (Diverse problems and inventions) of 1546, but this example was plagiarized from Chapter XV of Cardano's Ars magna of 1545, and he even introduced an error.

Cardano gave the solution in Latin:

RV:cubica 42 p:R 1700 p:RV:cubica 42 m:R  1700 m:2,

where V is the abbreviation of Vniversalis, p: means plus, and m: minus. Therefore, the root in modern notation is ${\sqrt[3]{42+\sqrt{1700}}}+{\sqrt[3]{42-\sqrt{1700}}}-2\,\,(=3.282260\dots).$

But Tartaglia's solution

R.u.cu.42.piu R.17000. piu R.u.cu.42.men.R.17000.men.2.

twice has 17000 instead of the correct 1700.

Scipione dal Ferro and Nicolo Tartaglia found the solution of the three cubic equations without a quadratic term.

But it was Cardano's great immortal feat to have solved the other ten cubic equations with a quadratic term, and thus to lead the way to a general solution.