Originally posted by kedaman
Yeah, newton raphson it is, dunno if it would get any faster with ASM

ok note I fixed you something, although I haven't tested it yet:
Code:
float root(float N){
float x,i;
 do{
    x=i;
    i=0.5*( x + N/x );
 }while(abs(i-x) > 0.00001);
 return i;
}

OK...I am testing this now....havn't found any info at the forum about QuerreryPerformance API, so I am trying to do this with GetTickCount. But I have more wuestions here.

N is the value you want to find the square root of.

i is after a while the square root. But what is x? The last iteration of the formula? So that should start as N?