solve square root equation x=6[squareroot(x-2)-1]?

We'll impose the constraint of existence of the square
root:

x-2>=0

We'll add
2 both sides:

x>=2

The
interval of admissible values is [2 ; +infinite).

We'll
re-write the given equation. We'll isolate 6sqrt(x-2) to the right side. For this
reason, we'll add 6 both sides:

6+x =
6sqrt(x-2)

We'll square raise both sides. For squaring the
binomial from the left side, we'll use the formula;

(a+b)^2
= a^2 + 2ab + b^2

36 + 12x + x^2 =
36(x-2)

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

We'll subtract
36x-72:

x^2 - 24x + 108 =
0

We'll apply the quadratic
formula:

x1=[24+ sqrt
(24^2-4*108)]/2

x1 =
(24+12)/2

x1 =
36/2

x1=18

x2=(24-12)/2

x2=6

Since
both values belong to the interval [2 ; +infinite), we'll validate them as solutions of
the given equation: x1 = 18 and x2 = 6.

Notes