how to solve the equation

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how to solve the equation
((3x+4)^(1/2))-((x+5)^(1/2))=1

(3x + 4)^(1/2) - (x + 5)^(1/2) = 1
(3x + 4)^(1/2) = (x + 5)^(1/2) + 1        [I squaring

3x + 4 = (x + 5) + 2(x + 5)^(1/2) + 1

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

x - 1 = (x + 5)^(1/2)                            [I squaring

x² - 2x + 1 = x + 5

x² - 3x - 4 = 0

x_1,2 = 1.5 ±√(2,25 + 4)

x₁ = 4

x₂ = -1

Greeting asinus :- )

  !

The square root of 4 is +2 not -2.

The roots of x^2 = 4 are +sqrt(4)and -sqrt(4), but sqrt(4) = +2.

Solve for x:
sqrt(3 x+4)-sqrt(x+5) = 1

(sqrt(3 x+4)-sqrt(x+5))^2 = 9+4 x-2 sqrt(x+5) sqrt(3 x+4) = 9+4 x-2 sqrt((x+5) (3 x+4)) = 1:
9+4 x-2 sqrt((x+5) (3 x+4)) = 1

Subtract 4 x+9 from both sides:
-2 sqrt((x+5) (3 x+4)) = -8-4 x

Raise both sides to the power of two:
4 (x+5) (3 x+4) = (-8-4 x)^2

Expand out terms of the left hand side:
12 x^2+76 x+80 = (-8-4 x)^2

Expand out terms of the right hand side:
12 x^2+76 x+80 = 16 x^2+64 x+64

Subtract 16 x^2+64 x+64 from both sides:
-4 x^2+12 x+16 = 0

The left hand side factors into a product with three terms:
-4 (x-4) (x+1) = 0

Divide both sides by -4:
(x-4) (x+1) = 0

Split into two equations:
x-4 = 0 or x+1 = 0

Add 4 to both sides:
x = 4 or x+1 = 0

Subtract 1 from both sides:
x = 4 or x = -1

sqrt(3 x+4)-sqrt(x+5) => sqrt(4+3 (-1))-sqrt(5-1)  =  -1:
So this solution is incorrect

sqrt(3 x+4)-sqrt(x+5) => sqrt(4+3 4)-sqrt(5+4)  =  1:
So this solution is correct

The solution is:
Answer: | x = 4