assuming your expression is this one:

we won't simplify even though we can
we can't take the square root of negative numbers nor can we divide by 0
so solve for the values that do that
ok, so the square root bits
we got √(28(x-1))
28(x-1)
this can't equal a negative
so when will it be negative?
28(x-1)<0
x-1<0
x<1
at x<1
so values less than 1 are restricted
other square root
√(8x²)
8x²<0
false
ok, and no dividing by 0
set √(8x²)=0
that is at x=0
so our restricted values are x<1 and x=0
so basically x<1 is all restricted values
so x≥1 is the set of all values that make it defined