1. Find the Real Solutions

I need to find the real solutions to a simple equation:

$\displaystyle \sqrt{x+5}=x$

$\displaystyle x+5=x^2$

$\displaystyle x^2-x-5=0$

$\displaystyle x=\frac{1\pm \sqrt{21}}{2}$

It seems that there should be two real solutions to the equation. However, I am told that the equation has only 1 real solution.

I need to find the real solutions to a simple equation:

$\displaystyle \sqrt{x+5}=x$

$\displaystyle x+5=x^2$

$\displaystyle x^2-x-5=0$

$\displaystyle x=\frac{1\pm \sqrt{21}}{2}$

It seems that there should be two real solutions to the equation. However, I am told that the equation has only 1 real solution.
From the first equation you know that $\displaystyle \sqrt{x+5}$ is a positive real number. Therefore x must be a positive real number.
But your last equation has also a negative number which can't be a solution of the original equation.

3. Only one solution

There is only one real solution: 2.79129.

In your first step, you squared both sides. You forgot that it is "plus or minus" though, when you do that. Make sense?