Solve for x sqrt(2x+5) = sqrt(x+2) + sqrt(2x-3).

Before solving the equation, we'll set the constraint of
existence of the sqrt.

2x + 5
>=0

2x>=-5

x>=-5/2

x+2>=0

x>=-2

2x-3>=0

2x>=3

x>=3/2

The
interval of admissible values for x is [3/2 , +inf.)

Now,
we'll solve the equation, raising to square both sides, in order to eliminate the square
root:

[sqrt(2x+5)]^2 = [sqrt(x+2) +
sqrt(2x-3)]^2

2x + 5 = x + 2 + 2x - 3 +
2sqrt(x+2)(2x-3)

We'll keep only the sqrt to the right
side, the rest of the terms we'll move them to the left
side:

2x + 5 -x - 2 - 2x + 3 =
2sqrt(x+2)(2x-3)

We'll eliminate like
terms:

6-x =
2sqrt(x+2)(2x-3)

We'll raise to square
again:

(6-x)^2 =
4*(x+2)(2x-3)

We'll expand the square and we'll remove the
brackets:

36 - 12x + x^2 = 8x^2 + 4x -
24

We'll move all terms to one
side:

36 - 12x + x^2 - 8x^2 - 4x + 24 =
0

We'll combine like
terms:

-7x^2 - 16x + 60 =
0

We'll multiply by -1:

7x^2 +
16x - 60 = 0

We'll apply the quadratic
formula:

x1 = (-16 + 44) /
14

x1 = 2

x2 = (-16 - 44) /
14

x2 = -60 / 14

x2 =
-30/7

Since the value of the first root
belongs to the interval of admissible values of x, the equation will have only one
solution, namely x = 2.