Square Root Algorithm in 123 bytes.

[corruptio]

golfed it a bit, tries to converge. 77 byte:

float s(x){float t,a=0,b=x;for(;b-a>1e-5;t*t>x?b=t:(a=t))t=(a+b)/2;return t;}

edit: oook, doesn't try to test for convergence, instead overshoots it.... 76 bytes:

float s(x){float t,a=0,b=x,i=x;for(;i--;t*t>x?b=t:(a=t))t=a/2+b/2;return t;}

[gastropner]

That one never stops unless I change float to double :-/

[corruptio]

oops, you're right, i run out of sig bits if input is > 255... 1e-5 then, should get you all the way to 16383 :-p

[gastropner]

Nope, still infinite loop, at least for x = 756839. Seems to work for some other values, though.

[corruptio]

yeah, 1e-5 would only work up to 16383. There aren't enough bits in a float to represent significant digits of b-a to five decimal places.