I am asked to solve this

$\displaystyle \frac{10}{3-4i}$

I looked over the book to see how this is done...

*according to the book* *$\displaystyle \frac{1}{3+4i}$ is to be written in stnrd form a+bi, so then one must multiply both the denominator and numerator by the opposite of the denominator in this case **3-4i*.
$\displaystyle \frac{1}{3+4i}.\frac{3-4i}{3-4i}$=$\displaystyle \frac{3-4i}{9+16}$

Ok here is where I don't get it, what happened(s) to the $\displaystyle i$ and the signs in the denominator of $\displaystyle \frac{3-4i}{9+16}$

is the multiplication process of the denominator the same as using foil?! or am I just full of complex numbers!?
$\displaystyle \frac{3-4i}{9+16}$ = $\displaystyle \frac{3}{25}-\frac{4}{25}$$\displaystyle i$

Where did the 25 come from?!
Thanks in advance to all!!