What is x if (x+1)^1/2=5-x ?

We'll impose the constraint of existence of the square
root:

x+1 >= 0

x
>= -1

Now, we'll solve the equation by raising the
square both sides:

[sqrt(x+1)]^2 =
(5-x)^2

We'll expand the square form the right
side:

x + 1 = 25 - 10x +
x^2

We'll subtract x+1 both
sides:

x^2 - 10x + 25 - x - 1 
=0

We'll combine like
terms:

x^2 - 11x + 24 =
0

We'll apply the quadratic
formula:

x1 = [11+sqrt(121 -
96)]/2

x1 =
(11+5)/2

x1 =
8

x2 =
3

Since both solutions are in
the interval of admissible values, they are
accepted.