Algebra mistake related to the equation $2\sqrt x=x-2$ The original equation is $2\sqrt x=x-2$ and I replaced $x$ with $4-2\sqrt3$. I am not sure what I did wrong with the algebra. Could someone please help me. My work is  posted below.
The problem is on the very first line:

$$2\sqrt{4-2\sqrt3} \stackrel{?}= 4-2\sqrt3-2.$$

The quantity on the left side of the equation is positive, but the quantity on the right is negative.
Therefore it is impossible for this equality to be true.

What the rest of the steps show is that the left-hand side and right-hand side are exactly opposite. (Different signs but same magnitude.)
That is, 
$$2\sqrt{4-2\sqrt3} = -(4-2\sqrt3-2).$$
Nothing is wrong (except I never saw squaring of equations like that).

You calculations show, that $x = 4 - 2\sqrt{3}$ is indeed the solution of your original equation.

Try other numbers. There is one more.