...something like this?
C1: f1(x):= x - 2 * sqrt (x);
C2: solve (f(x) = 10, x);
gives back
D2: [x = 2 SQRT(x) + 10]
which IMHO isn't much of a solution :-) But when I substitute y = sqrt
(x) to get an equivalent function
C3: f3(y):= y^2 - 2 * y;
C4: float (solve (f2(y) = 10, y);
I get two answers [y = -2.316, y = 4.316], and if I square the second, I
get a solution to f1?
I'm puzzled.
James