What is 12/(square root of 2 - 6)?

I'm not quite sure on your notation here, I'm assuming you're meaning this `12/(sqrt2 - 6)` and not `12/sqrt(2-6)`.

To do this problem we just need to rationalize. The concept in rationalizing is quite simple, we know that `(x-y)(x+y) = x² - y²`.

So to get rid of these roots on the denominator, we'll multiply it by `sqrt2 + 6`. Which is the same thing as the denominator but with the sign switched so we won't have any roots on the bottom to deal with.

But - and there's always a but - since this is a fraction I can't just multiply what's on the denominator. I need to multiply both the numerator and denominator by the same thing, so it goes:

`12/(sqrt2 - 6) = 12/(sqrt2 - 6) * (sqrt2 + 6)/(sqrt2 + 6)`
`12/(sqrt2 - 6) = 12 * (sqrt2 + 6)/((sqrt2)^2 - 6^2)`
`12/(sqrt2 - 6) = (12sqrt2 + 12*6)/(2 - 36)`

We can put a 2 on evidence both on the numerator and on the denominator

`12/(sqrt2 - 6) = (2*(6sqrt2 + 6*6))/(2*(1 - 18))`
`12/(sqrt2 - 6) = (6sqrt2 + 6*6)/( - 17)`

17 is a prime number so we don't really have much more to do here. You can either put that 6 on evidence on the numerator, or evaluate `6^2`

`12/(sqrt2 - 6) = -(6*(sqrt2 + 6))/(17)` or
`12/(sqrt2 - 6) = -(6sqrt2 + 36)/(17)`