(x+2)^1/2-(2-x)^1/2=2

First we will square both sides:

==> [(x+2)^1/2 - (2-x)^1/2] = 2^2

==> (x+2) - 2sqrt(x+2)(2-x) + (2-x) = 4

==> (x+2 + 2 -x) - 2sqrt(-x^2 - 4x + 4) = 4

==> 4 - 2sqrt(-x^2 -4x + 4) = 4

Now we will move 4 to the right side:

==> -2sqrt(-x^2 - 4x + 4) = 0

Divide by -2:

==> sqrt(-x^2 - 4x + 4) = 0

==> -x^2 - 4x + 4 = 0

Divide by -1:

==> x^2 + 4x - 4 = 0

==> x1= (-4+sqrt(16+16)/ 2 = (-4+ 4sqrt2)/2 = -2+2sqrt2

==> x2= -2-2sqrt2 ( trhis solution is impossible because the function is not defined for x2 value.

**Then the solution is:**

**x = { -2+2sqrt2 }**

First, we'll impose the constraint of existence of the square roots:

x+2 >=0

x> =-2

2-x>=0

x=<2

The interval of admissible values for x is [-2 ; 2].

Now, we'll solve the equation.

We'll multiply both sides by the conjugate of the expression sqrt(x+2) +sqrt(2-x).

sqrt(x+2) - sqrt(2-x) = 2 (1)

x + 2 - 2 + x = 2[sqrt(x+2) +sqrt(2-x)]

We'll combine and eliminate like terms:

2x = 2[sqrt(x+2) +sqrt(2-x)]

We'll divide by 2:

[sqrt(x+2) +sqrt(2-x)] = x (2)

We'll add (1) + (2):

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

We'll combine and eliminate like terms:

2sqrt(x+2) = x+2

We'll square raise both sides:

4(x+2) = (x+2)^2

We'll subtract 4(x+2) both sides:

(x+2)^2 - 4(x+2) = 0

We'll factorize by (x+2):

(x+2)(x+2-4) = 0

We'll put each factor as zero:

x + 2 = 0

**x = -2**

x - 2 = 0

**x = 2**

**Since the values for the roots are the endpoints of the interval of admissible values, we'll accept both.**