Answer:
(-infinity, 9/8]
Step-by-step explanation:
Domain is defined as all the possible x-values of a function.

Lets test some negative numbers.


OK! That means all negative numbers should work!
What about 0?

0 works..
Now, onto positive numbers?

=> imaginary
Check the last value that works. It can only be 9/8 because that's where the square root becomes 0, which is the last value before you head into the imaginary numbers..

(-infinity, 9/8]