How do you factor this equation that has square roots involved? I'm trying to solve this equation for 
$$\sqrt {16+l^2} + \sqrt {4+l^2} = {nv \over f}$$ 
I know what ${nv \over f}$ is, but no matter what I do I keep getting a weird answer. I've resulted to using Wolfram Alpha, but I'd really like to know how to do this myself. 
 A: Square-rearrange-square again as garserdt216 outlined totally works. I'm not going to solve your problem but here see it applied to an equivalent generalized problem
$$\sqrt{x + a} + \sqrt{x + b} = c$$
Square
$$(x + a) + 2\sqrt{(x + a)(x + b)}+ (x + b) = c^2$$
Isolate square root
$$\sqrt{(x + a)(x + b)} = \frac{c^2 - a - b}{2} - x$$
Square again
$$(x + a)(x + b) = \left (\frac{c^2 - a - b}{2} - x \right )^2$$
Simplify
$$x^2 + (a + b)x + ab = x^2  + (a + b - c^2)x+ \left (\frac{c^2 - a - b}{2}\right)^2 $$
Simplify some more
$$x = \frac{1}{c^2}\left ( \left (\frac{c^2 - a - b}{2}\right)^2 - ab \right ) = \frac{(c^2 - a - b)^2 - 4ab}{4c^2}$$
(Check for consistency with original equation.)
EDIT: Thanks to @Claude Leibovici for pointing out some errors which I've now hopefully corrected. 
A: Hint: let $4+l^2=t^2 \gt 0\,$, assume WLOG $t \gt 0\,$, then write the equation as: $$t + \sqrt{t^2+12} = \frac{nv}{f}\quad\implies\quad t^2+12 = \left(\frac{nv}{f} - t\right)^2$$
Solve for $\,t\,$, then solve $\,l^2=t^2-4$ for $\,l\,$ and verify the solution(s) since the derivation used one-way implications, not equivalencies.

[ EDIT ]   In reply to a comment, the final verification can indeed be avoided if, instead, the assumption $t \gt 0$ is enforced upon the solutions in $t\,$. In other words, the last line above can be replaced with: "Solve for $\,t\,$, drop any non-positive solutions, then solve $\,l^2=t^2-4$ for $\,l\,$".
A: I may have done some mistakes in my calculations since, using the same steps as Squid's answer, I end with $$x=\frac{\left(c^2-a-b\right)^2-4 a b}{4 c^2}$$ So, if $a=16$, $b=4$,$\frac{nv}f=c$, this would lead to $$l^2=\frac{c^2}{4}+\frac{36}{c^2}-10=\frac{1}{4} \left(c+\frac{12}{c}\right)^2-16$$
A: Squaring both sides (and not bothering absolute values for now) gives
$$\left(16+l^2\right)+2\sqrt{\left(16+l^2\right)\left(4+l^2\right)}+\left(4+l^2\right)=\frac{n^2\nu^2}{f^2}.$$
Now after isolating the square root and squaring again
$$
\left(\sqrt{\left(16+l^2\right)\left(4+l^2\right)}\right)^2=\left(\frac{\frac{n^2\nu^2}{f^2}-\left(16+l^2\right)-\left(4+l^2\right)}{2}\right)^2.$$
So now,
$$64+20l^2+l^4=\frac{1}{4}\left(\frac{n^2\nu^2}{f^2}-20-2l^2\right)^2=\frac{1}{4}\left(\frac{n^4\nu^4}{f^4}+400+4l^4-\frac{40n^2\nu^2}{f^2}+80l^2-\frac{4l^2n^2\nu^2}{f^2}\right)$$
Rewrite this in the form
$$Al^4+Bl^2+C=0$$
and let $x=l^2$. Solve this quadratic polynomial in $x,$ square it and you'll get $l.$ Go back and repeat this but use an absolute value whenever you have the square of a square root because there might be other solutions.
