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secret, in order to secure by that means an advantage over rivals by proposing problems beyond their reach. This practice gave rise to numberless disputes regarding the priority of inventions. A second solution of cubics was given by Nicolo of Brescia (1506(?)-1657). When a boy of six, Nicolo was so badly cut by a French soldier that he never again gained the free use of his tongue. Hence he was called Tartaglia, i.e. the stammerer. His widowed mother being too poor to pay his tuition in school, he learned to read and picked up a knowledge of Latin, Greek, and mathematics by himself. Possessing a mind of extraordinary power, he was able to appear as teacher of mathematics at an early age. In 1530, one Colla proposed him several problems, one leading to the equation $\scriptstyle {x^{3}+px^{2}=q}$. Tartaglia found an imperfect method for solving this, but kept it secret. He spoke about his secret in public and thus caused Ferro's pupil, Floridas, to proclaim his own knowledge of the form $\scriptstyle {x^{3}+mx=n}$. Tartaglia, believing him to be a mediocrist and braggart, challenged him to a public discussion, to take place on the 22d of February, 1535. Hearing, meanwhile, that his rival had gotten the method from a deceased master, and fearing that he would be beaten in the contest, Tartaglia put in all the zeal, industry, and skill to find the rule for the equations, and he succeeded in it ten days before the appointed date, as he himself modestly says.^[7] The most difficult step was, no doubt, the passing from quadratic irrationals, used in operating from time of old, to cubic irrationals. Placing $\scriptstyle {x={\sqrt[{3}]{t}}-{\sqrt[{3}]{u}}}$, Tartaglia perceived that the irrationals disappeared from the equation $\scriptstyle {x^{3}+mx=n}$, making $\scriptstyle {n=t-u}$. But this last equality, together with $\scriptstyle {({\frac {1}{3}}m)^{3}=tu}$, gives at once

$\scriptstyle {t={\sqrt {({\frac {n}{2}})^{3}+({\frac {m}{3}})^{3}}}+{\frac {n}{2}},~u={\sqrt {({\frac {n}{2}})^{2}+({\frac {m}{2}})^{3}}}-{\frac {n}{2}}}$.