We have to solve (2x+5)^1/2-(x-2)^1/2=3

(2x+5)^1/2-(x-2)^1/2=3

square both the sides

2x + 5 + x - 2 - 2*(2x+5)^1/2*(x-2)^1/2 = 9

=> 3x + 3 - 9 = 2*(2x+5)^1/2*(x-2)^1/2

=> 3x - 6 = 2*(2x+5)^1/2*(x-2)^1/2

square the sides again

=> 9x^2 + 36 - 36x = 4*(2x+5)(x-2)

=> 9x^2 + 36 - 36x = 4*(2x^2 - 4x + 5x - 10)

=> 9x^2 + 36 - 36x = (8x^2 - 16x + 20x - 40)

=> x^2 - 40x + 76 = 0

=> x^2 - 38x - 2x + 76 = 0

=> x(x - 38) -2(x - 38) = 0

=> (x - 2)(x - 38) = 0

=> x = 2 and x = 38

**The solution for x is (2 , 38)**

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

We'll add sqrt(x-2) both sides:

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

We'll raise to square both sides:

2x + 5 = 9 + 6sqrt(x-2) + x - 2

We'll subtract x - 2:

x + 7 - 9 = 6sqrt(x-2)

x - 2 = 6sqrt(x-2)

We'll raise to square again:

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

We'll substitute x - 2 = t:

t^2 - 36t = 0

We'll factorize by t:

t(t - 36) = 0

t = 0

t = 36

x - 2 = 0

**x = 2**

x - 2 = 36

**x = 38**

**Since x has to be larger or equal to 2, both solutions will be accepted.**