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sage: var('u')
u
sage: solve(u==x^2+1, x)
[x == -sqrt(u - 1), x == sqrt(u - 1)]

So you could use [1] instead of [0] to get the "positive" one.

I don't know how you got the r1 from this example, but the documentation for solve says, among other things:

   If there is a parameter in the answer, that will show up as a new
   variable.  In the following example, "r1" is a real free variable
   (because of the "r"):

      sage: solve([x+y == 3, 2*x+2*y == 6],x,y)
      [[x == -r1 + 3, y == r1]]

Hope this helps!