ah ok
in mathomatic I got:
1-> y=a^2+a/3
a
#1: y = (a^2) + -
3
1-> a
Equation is a degree 2 polynomial in (a).
Equation was solved with the quadratic formula.
1
-(1 + (((1 + (36*y))^-)*sign))
2
#1: a = ------------------------------
6
I thought there was a way to get the same format also in maxima
p.s.: another question:
If I write: ax+ay+az;
how to get: a(x+y+z) ?
many thanks
--- Gio 22/11/12, Evgeniy Maevskiy <emaevskiy at e-math.ru> ha scritto:
> Da: Evgeniy Maevskiy <emaevskiy at e-math.ru>
> Oggetto: Re: [Maxima] how to calculate "a" from this equation?
> A: "Antonio Lapira" <antoniolapira at yahoo.it>
> Data: Gioved? 22 novembre 2012, 16:34
> eq: y=a^2+a/3;
> solve(eq,a);
>
>
> 22.11.2012 16:39, Antonio Lapira ?????:
> > if I enter: y=a^2+a/3;
> > then how can I see the formula a= ...... ?
> > (not the result, but the formula)
> >
> > thanks
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> >
>
>