What are the real solutions of equation square root (x+1) + square root( 2x+3) = 5 ?

before solving the equation, we'll impose the constraints
of existence of square root:

x + 1 >= 0 => x
>= -1

2x+ 3 >=
0

2x >= -3

x>=
-3/2

The common interval of admissible values of x is [-1 ,
+infinite).

We'll solve the equation by raising to square
both sides:

x+1+2x+3 + 2sqrt(x+1)(2x+3) =
25

3x + 4 + 2sqrt(x+1)(2x+3) =
25

2sqrt(x+1)(2x+3) = 21 -
3x:

2sqrt(x+1)(2x+3) = 3(7 -
x)

We'll raise to square
again:

4(x+1)(2x+3) =
9(7-x)^2

We'll expand the square and w'ell remove the
brackets:

8x^2 + 20x + 12 = 441 - 126x +
9x^2

x^2 - 126x - 20x + 441 - 12 =
0

x^2 - 146x + 429 = 0

x1 =
[146 + sqrt(21316 - 1716)]/2

x1 = (146 +
140)/2

x1 = 143

x2 = (146 -
140)/2

x2 =
3

Since both values of x are in the common
interval [-1 , +infinite), we'll validate them as solutions: {3 ;
143}.