You are very welcome Andrey,
I am glad that I found it. On the other hand I would be very grateful to you if this would work with either g(x) or m(x)
solve(g(x),x,1,2)=
roots(g(x),x,1)=
giving the result. It is very important that root finding functions have the possibility to accept this.
Regards,
Radovan
When Sisyphus climbed to the top of a hill, they said: "Wrong boulder!"