Question 631354
I assume that a pencil and paper calculation is required, because a calculator would give you the answer easily.
  
If the square roots are going to be rational numbers, the way to go is:
{{{sqrt(640.09)=sqrt(64009/100)=sqrt(64009)/sqrt(100)=sqrt(64009)/10}}} and
{{{sqrt(9682.56)=sqrt(968256/100)=sqrt(968256)/sqrt(100)=sqrt(968256)/10}}}
If the square roots are going to be rational numbers, 64009 and 968256 must be perfect squares.
From there, I would try to factor the numbers in the square roots.
A complete prime factorization is not needed, but the idea is similar.
 
64009 is not divisible by 2 (it is not even) or 3 (digits do not add to a multiple of 3).
It is not divisible by 5 (does not end in 0 or 5) or by 7 (I tried dividing).
It is divisible by 11, because the sums of alternate digits differ by 11:
(6+0+9)-(4+0)=15-4=11
If is is a perfect square, {{{11^2}}} will be a factor, so I can divide by 11 twice.
I did, and I got 64009/11=5819 and 5819/11=529.
The number 529 sounded familiar, as if I had known it as a perfect square.
Since 529 ends in 9, it must be the square of a number that ends in 3 or 7.
That number be smaller than 25, because {{{25^2=625>529}}} ,
but larger than 20 because {{{20^2=400<529}}}
I tried 23 and found that {{{23^2=529}}}
So {{{64009=11*11*23^2=11^2*23^2=(11*23)^2=253^2}}} and {{{sqrt(64009)=253}}}
So {{{sqrt(640.09)=sqrt(64009)/10=253/10=25.3}}}
 
968256 is divisible by 2, and if it is a perfect square it must be divisible by {{{2^2=4}}}.
So I divided by 4 once to get 968256/4=242064, and again to get 242064/4=60516,
and a third time to get 60516/4=15129, that is not divisible by 2 any more.
15129 is divisible by {{{3^2=9}}} and I divided 15129/9=1681.
If 1681 going to be a perfect square, it must be the square of a number that ends in 1 or 9, and is a bit larger than 40, because {{{40^2=1600<1681}}}.
I tried 41 and found that {{{41^2=1681}}} so
41^2*9*4*4*4=41^2*3^2*64=41^2*3^2*8^2=(41*3*8)^2=984^2=968256 and {{{sqrt(968256)=984}}}
So {{{sqrt(9682.56)=sqrt(968256)/10=984/10=98.4}}}
 
There is a general way to find square roots, that even allows you to get approximate values for irrational squre roots. It can be worked into a procedure sort of like long division, which I was taught in school many, many years ago. I hope that is not what was expected from you in the era of smartphones and tablet computers.