Re: Solutions to a cubic?

["Gerry"]

Gerry wrote:
> This one should work out just fine:
> ((3n+1)a^2+(3m+1)a+1)/a^2=3y1^2+3y1y2+y2^2
Let me try this again:

x1=-3,x2=-2,x3=0
a=-1/4
n=2, m=1

b=na+m=1/2
y1=x2-x1=-1
y2=x3-2x2+x1=1

(((3n+1)a^2+(3m+1)a+1)/a^2=7
3y1^2+3y1y2+y2^2=7

> Regards
> Gerry