When we simplify radicals in a fraction, we have to rationalize the denominator where the radical is at.

There is a rule that shows #(a-b)(a+b)=a^2-b^2#.

So in the fraction, #5/(sqrt3 - 2)# , the denominator #(sqrt3-2)# represents (a-b) in the rule of #(a-b)(a+b)=a^2-b^2#. To get rid of a surd, we can square the surd as seen in #a^2-b^2#

Let's place #sqrt3# as #a# and #2# as #b#.

Hence, #(sqrt3-2)##(sqrt3+2)# = #(sqrt3)^2#- #2^2#

As we remember, the square of a radical/surd rationalises the radical.

#(sqrt3)^2#= #3#

So what we can do is to muiltiply #5/(sqrt3 - 2)# x #(sqrt3+2)/(sqrt3 + 2)# where #(sqrt3+2)/(sqrt3 + 2)#= #1# so it does not make a difference to the value of the product.

#(5(sqrt3+2))/((sqrt3 - 2)(sqrt3+2)#=#(5sqrt3+10)/((sqrt3)^2-2^2)#=#(5sqrt3+10)/(3-4)#=#(5sqrt3+10)/(-1)#= # -5sqrt3-10 #

#5/(sqrt3 - 2)#=# -5sqrt3-10 #=#-18.66 (2 d.p.)#