Question 309970
If we let {{{x=2*sqrt(3)}}} and {{{y=sqrt(7)}}} we get the expression {{{(x-y)(x+y)}}}. FOILing this expression gives us {{{(x-y)(x+y)=x^2-y^2}}} which is a difference of squares. Plug {{{x=2*sqrt(3)}}} and {{{y=sqrt(7)}}} back in to get 


{{{(2*sqrt(3)-sqrt(7))(2*sqrt(3)+sqrt(7))=(2*sqrt(3))^2-(sqrt(7))^2}}} 



Now just multiply out and simplify the right side to get: 



{{{(2*sqrt(3)-sqrt(7))(2*sqrt(3)+sqrt(7))=4*3-7}}} 



and do a bit more arithmetic to get



{{{(2*sqrt(3)-sqrt(7))(2*sqrt(3)+sqrt(7))=5}}}