What are the real solutions of square root(x^2+9)=x-2 ?

Since the radicand (x^2 + 9) is positive for any real
value of x, we don't have to impose the constraint of existence of the square
root.

We'll raise to square both sides, to eliminate the
square root:

x^2 + 9 =
(x-2)^2

We'll expand the square from the right, using the
formula:

(a-b)^2 = a^2 - 2ab +
b^2

(x-2)^2 = x^2 - 4x + 4

The
equation will become:

x^2 + 9 = x^2 - 4x +
4

We'll subtract both sides x^2 - 4x +
4:

x^2 + 9 - x^2 + 4x - 4 =
0

We'll eliminate like
terms:

4x + 5 = 0

We'll shift
5 to the right:

4x = -5

x =
-5/4

The equation has only a real solution
and this is x = -5/4.