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If you look at the documentation of solve function:

sage: solve?

you will see e few options, in particular to_poly_solve.

If you do :

sage: S = solve(UU_I==0, I, to_poly_solve=True) ; S
[I == (7/3)]

Now, to extract the 7/3 value, you should beter get a list of dictionnaries than a list of expressions:

sage: S = solve(UU_I==0, I, to_poly_solve=True, solution_dict=True) ; S 
[{I: 7/3}]

Now it is easy to extract the solution:

sage: S[0][I]

7/3

And to evaluate UU on this value :

sage: UU(S[0][I])
2.14087209644419

Which looks consistent with the plot:

sage: plot(UU,I,-10,10)

If you look at the documentation of solve function:

sage: solve?

you will see e few options, in particular to_poly_solve.

If you do :

sage: S = solve(UU_I==0, I, to_poly_solve=True) ; S
[I == (7/3)]

Now, to extract the 7/3 value, you should beter get a list of dictionnaries than a list of expressions:

sage: S = solve(UU_I==0, I, to_poly_solve=True, solution_dict=True) ; S 
[{I: 7/3}]

Now it is easy to extract the solution:

sage: S[0][I]
7/3

And to evaluate UU on this value :

sage: UU(S[0][I])
2.14087209644419

Which looks consistent with the plot:

sage: plot(UU,I,-10,10)

If you look at the documentation of solve function:

sage: solve?

you will see e few options, in particular to_poly_solve.

If you do :

sage: S = solve(UU_I==0, I, to_poly_solve=True) ; S
[I == (7/3)]

Now, to extract the 7/3 value, you should beter get a list of dictionnaries than a list of expressions:

sage: S = solve(UU_I==0, I, to_poly_solve=True, solution_dict=True) ; S 
[{I: 7/3}]

Now it is easy to extract the solution:

sage: S[0][I]
7/3

And to evaluate UU on this value :

sage: UU(S[0][I])
2.14087209644419

Which looks consistent with the plot:

sage: plot(UU,I,-10,10)