What is now called ``Cardan's method" to solve the cubic, but that should be more correctly called ``del Ferro's method" goes thus: consider the general equation of third degree . A substitution gathers . Let . This implies that solving the above cubic is equivalent to solving the cubic where and . Now substitute t=u+v. This means that . The latter is satisfied when and 3uv+p=0. Hence and , so and are the roots of the quadratic . Hence one can find u and v and therefore t=u+v in terms of p and q. The quartic had been solved with a method not much different from the one above.