From the binomial theorem, you know that $(a + b)^2 = a^2 + 2ab + b^2$.
Consider the particular case where $a = -2$ and $b = \sqrt{x-4}$.
Then the theorem says that
$$(-2 + \sqrt{x-4})^2 = (-2)^2 + 2(-2)\sqrt{x-4} + (\sqrt{x-4})^2.$$
Simplify:
$$(-2)^2 + 2(-2)\sqrt{x-4} + (\sqrt{x-4})^2
= 4 - 4\sqrt{x-4} + |x - 4|.$$
Now if $x < 4$ then $\sqrt{x-4}$ is not a real number.
So we can assume $x \geq 4$, so $x - 4 > 0$,
and therefore $|x - 4| = x - 4$. So
$$ 4 - 4\sqrt{x-4} + |x - 4| = 4 - 4\sqrt{x-4} + (x - 4)
= x - 4\sqrt{x-4}.$$
Together with the equation you already had, this tells us that
$$ x + 8 = x - 4\sqrt{x-4}.$$
I think you can see how to simplify this further, but when you do you will
find that it is a false equation for every possible value of $x$.
Hence the equation you started with was false for every value of $x$.
In fact we can come to this conclusion even easier.
The square root is an increasing function, so from
$8 > -4$ we know that $x + 8 > x - 4$ and therefore
$$\sqrt{x + 8} > \sqrt{x - 4}.$$
That implies that
$$\sqrt{x + 8} - \sqrt{x - 4} > 0,$$
that is, it is impossible for $\sqrt{x + 8} - \sqrt{x - 4}$ to be negative,
and in particular it is impossible for it to be $-2$.
There is no need to square anything and no need for the binomial theorem.