Why isn't this root equation working out?

Question

Suppose we have the equation

$\sqrt{3x-2}=\sqrt{x-2}+2$

I know the solutions are 2 and 6 but I can't figure out where I'm going wrong in the following

$(\sqrt{3x-2})^{2}=(\sqrt{x-2}+2)^{2}$

$3x-2=x-2+4\sqrt{x-2}+4$

$2x-4=4\sqrt{x-2}$

$4x^2-32x+48=0$

But this gives imaginary roots

If I rearrange it so that we have

$\sqrt{3x-2}-2=\sqrt{x-2}$

I now get real roots but they're still not 2 and 6

My question is two fold

1. how do I do this to obtain the proper solutions of two and six

2. why does rearranging it lead to different solutions, some real and some imaginary ?

Answer 1

$$4x^2-32x+48 = 0$$ can first be simplified to

$$x^2-8x+12 = 0$$

and this equation has a discriminant of $$b^2-4ac = (-8)^2-4\cdot 1 \cdot 12 = 64-48=16>0$$ which means the equation has real solutions.