How to solve this proportion Okay I think I'm just having a major brain block, but I need help solving this proportion for my physics class.
$$\frac {6.0\times 10^{-6}}{ x^2} = \frac {2.0\times 10^{-6}}{ (x-20)^2}$$
What's confusing me is the solution manual to this problem lists writing the proportion as, 
$$\frac {(x-20)^2} { x^2} = \frac {2.0\times 10^{-6}}{ 6.0\times 10^{-6}}$$
and then proceeds to solve the problem from there... but that doesn't seem right to me. Usually you would cross multiply a proportion and solve, but they seemed to do some illegal math or something. Could you guys work me through how to solve this? This answer is 47 by the way.
 A: It's only $47$ in physics.  In mathematics, it would be
$$
\frac{20\sqrt{3}}{\sqrt{3}-1} = 47.320\ldots
$$
:-)

Anyway: You start off with a proportion that can be written, generally, as
$$
\frac{a}{b} = \frac{c}{d}
$$
The book then proceeds to rewrite this as
$$
\frac{d}{b} = \frac{c}{a}
$$
That the two are equivalent (provided $a \not= 0$) can be seen by multiplying both sides of the first equation by $d$, and then dividing by $a$.  You can also see that both equations yield the same result after cross-multiplication.
A: The proportion they gave is correct. Simply divide both sides of the equation you started with by $6\cdot10^{-6}$ and multiply both sides by $(x-20)^2$. 
So we have 
$$
\frac{(x-20)^2}{x^2} = 
\frac{2.0\cdot10^{-6}}{6.0\cdot10^{-6}}=\frac{1}{3}.$$
The left-hand side becomes 
$$\frac{(x-20)^2}{x^2}=\frac{x^2-40x+400}{x^2}=\frac{1}{3}.$$
Multiplying both sides by $x^2$, we get
$$x^2-40x+400=\frac{x^2}{3} \quad\Rightarrow\quad\frac{2}{3}x^2-40x+400=0.$$
From here you use the quadratic equation to give you $x=10(3±\sqrt{3})$. It's physics, so you probably want the positive one, so $x=10(3+\sqrt{3})\approx47$. 
