Solve for real x the equation x=6[(x-2)^1/2-1].

For the beggining, we'll impose constraint of existence of
the square root:

x - 2 >=
0

x >= 2

The real
solutions of the equation have to belong to the range [2 ;
+infinite).

Now, we'll solve the equation. We'll start by
removing the brackets:

x = 6sqrt(x-2) -
6

We'll add 6 both sides, in order to isolate the square
root to the right side.

x + 6 =
6sqrt(x-2)

Now, we'll square raise to eliminate the square
root;

x^2 + 12x + 36 = 36x -
72

We'll subtract 36x -
72:

x^2 + 12x + 36 - 36x + 72 =
0

We'll combine like
terms:

x^2 - 24x + 108 =
0

We'll apply quadratic
formula:

x1 =[24 + sqrt(576 -
432)]/2

x1 = (24 + 12)/2

x1 =
18

x2 = (24 - 12)/2

x2 =
6

Since both values belong to the interval of
admissible values, we'll accept them. The solutions of the equation are: {6 ;
18}.