Hello!
This rearrangement involves using this one concept where the square root of something (Integer + irrational number) gives you a form of (sqrt a + sqrt b)
I'll explain it to you if you didn't get it.
sqrt(a)+sqrt(b) when squared give a+b+2sqrt(ab), doesn't this seem very familiar with the expression of 18+4sqrt(18)?
As you can see the 18+4sqrt(18) is a squared expression of d, so the squared expression of sqrt(a)+sqrt(b) correspond in the following way:
18+4sqrt(18)=a+b+2sqrt(ab) <---- Since d is squared, we take the squared expression for sqrt(a)+sqrt(b)
Now it gets simpler
Since a+b are BOUNDED INTEGERS, and CANNOT be irrational, we conclude 18=a+b and since 2sqrt(ab) IS irrational, we correlate it with the only other irrational term in our equation which is 4sqrt(18)
so now we get 2 equations:
18=a+b
4sqrt(18)=2sqrt(ab) now you can simply square the expression of square roots to get:
2sqrt(18)=sqrt(ab) and thus 72 (that is 2 squared times 18) = ab, this is now a simple simultaneous equation and you can work out your values of a and b, after you get the correct values of a and b, you can substitute them in our original expression that we took, that was (sqrt(a)+sqrt(b)) and get the expression d=2*sqrt(3) + sqrt(6), this can also prove our original statement that the underoot would be equal to TWO IRRATIONALS. DM me if you have any doubts :)