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)#