Solve for x the equation x+2=squareroot(2x+12)?

The first step is to impose constraint of existence of the
radical.

2x+12 `>=`
0

We'll isolate 2x to the left
side:

2x `>=` -12

x
`>=` -6

The values of x that can be validated as
solutions of the given equation must belong to the closed interval [-6,+`oo`
).

Now, we'll solve the equation raising to square both
sides, to remove the radical:

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

We'll expand the
binomial:

x^2 + 4x + 4 = 2x +
12

We'll shift all terms to the left
side:

x^2 + 2x - 8 = 0

We'll
apply quadratic formula:

x1 = [-b+sqrt(b^2 -
4ac)]/2a

a=1 , b = 2 and c =
-8

x1 = [-2+sqrt(4 + 32)]/2

x1
= (-2+6)/2

x1 = 2

x2 =
(-2-6)/2

x2 =
-4

Since both values of x are located in the
closed interval [-6 , `oo` ), the solutions of the equation are {-4 ;
2}.