What is the value of x in 2x+12*[1 - (x-2)^(1/2)]=0?

Before solving a square root equation, we'll have to
impose the constraint of existence of the square root.

The
radicand has to be positive:

x -
2>=0

x>=2

So,
all the solutions of the equation have to belong to the interval
[2;+infinite).

Now, we'll solve the equation. We'll divide
by 2:

x + 6 - 6sqrt(x-2) =
0

We'll move - 6sqrt(x-2) to the right side, so that
raising to square both sides, we'll eliminate the square
root.

(x+6)^2 = [6square
root(x-2)]^2

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

We'll remove the
brackets:

x^2 + 12x + 36 - 36x + 72 =
0

We'll combine like
terms:

x^2 - 24x + 108 = 0

x1
= [24+sqrt(144)]/2

x1 =
(24+12)/2

x1 = 18

x2 =
6

Since both values are in the interval of
possible values, we'll validate them as solutions: x1 = 18 and x2 =
6.