Square root -x+6=x-6 identify any extraneous solution. x=

I don't quite see any extraneous solutions, other than, the root is an even root, root of 2, and therefore, it will give you a ± solutions,  if x = 6, the radicand becomes 0 then, and the roots for that will be ±0, keeping in mind 0 is neither per se.

now on the right-hand-side, 6-6 is just a flat positive 0, so the +0 root will match that, however the -0 will not, and thus can be seen as extraneous.