Help with easy quadratic

The question is:

Solve (x-8)^2 = the square root of 11

It's completing the square for sure and I've tried a number of different ways each resulting in different answers.
I've tried bringing the square root of 11 over and then solving it, I've tried halfing 8 and then having -16 outside the bracket and then solving it and I've tried solving it straight away from what I'm given.

If anyone could point me in the right direction that'd be helpful.

Square root both sides to get
x-8 = cubed root of 11
so its
x = 8 + cubed root of 11
That might be all you have to write but if they want it as a decimal, you will have to use a calculator to find out the cubed root of 11.
8 + 2.22 (because its a cubed root you only do the + and not ±)
so x = 10.22

Hope that helps you.

Thank you. I got an answer very similar to this, but it seemed a bit too straightforward, as silly as that sounds.

I was thinking it may be wanting me to solve it as if it was in the form x^2 + 8x.
What do you think of this?

You square root both sides.
So you get x-8=11^0.25
Then x=11^0.25 +8
Work that out n U get the answer, that's all.
Btw the guy above was wrong he said cubed root.
You would then get two answers. Because 11^0.25 can be positive or negative

Alternatively you could expand the bracket in the left; x^2-16x+64-11^0.5=0 and then to the quadratic formula and you would get the right answers.