In this case you do not want a negative argument for the square root (you cannot find the solution of a negative square root, at least as a real number).

What you do it is to “impose” that the argument is always positive or zero (you know the square root of a positive number or zero).

So you set the argument bigger or equal to zero and solve for ##x## to find the ALLOWED values of your variable:

##6-2x>=0## ##2x<=6## here I changed sign (and reversed the inequality).

And finally: ##x<=3##

So the values of ##x## that you can accept (domain) for your function are all the values smaller than ##3## including ##3##. Check by yourself substituting for example ##3##, ##4## and ##2## to confirm our deduction.