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.