How to do this division involving radicals, by hand? Is it possible to perform this division by hand?
$$ x= \dfrac{2\sqrt{6} - \sqrt{30} + 14\sqrt{5} - 43}{2\sqrt{30} + 2 - 3\sqrt{6}-3\sqrt{5}} = 4 + \sqrt{5} + 2\sqrt{6} $$
If so, how? And is this possible for any general division involving multiple radicals, or this one is possible just because of its beautiful simplification?
I suspect that the $\sqrt{30}$ is of some significance, but I don't know how. I verified this here: Wolfram Alpha result. For those who are curious, it is the $x$-coordinate of the orthocentre of triangle $ABC$ where $A=(2\sqrt 6,1),\;B=(4,3),\;C=(\sqrt 5, 2\sqrt 5)$. Thankfully, it was I multiple choice test and I just objective'd it, but I just got curious, so I asked.
 A: If your denominator has $4$ or fewer square root terms, you can "rationalize" using $$\begin{align*}
& (a\sqrt{A}+b\sqrt{B}+c\sqrt{C}+d\sqrt{D})(a\sqrt{A}+b\sqrt{B}-c\sqrt{C}-d\sqrt{D})\\
&=a^2A+b^2B+2ab\sqrt{AB}-c^2C-d^2D-2cd\sqrt{CD}
\end{align*}$$
to reduce it to $2$ square root terms. Now the denominator would have the form $x+y\sqrt{Y}+z\sqrt{Z}$, which you can "rationalize" using
$$(x+y\sqrt{Y}+z\sqrt{Z})(x+y\sqrt{Y}-z\sqrt{Z})=x^2+y^2Y+2xy\sqrt{Y}-z^2Z.$$
Then the denominator has the form $u+v\sqrt{V}$, which you "rationalize" as 
$$(u+v\sqrt{V})(u-v\sqrt{V})=u^2-v^2V.$$
Now you have an integer denominator, but what about the numerator?
Because at the beginning you have only $5$, $6$, and $30$ in radicals, and because $\sqrt{5}\sqrt{6}=\sqrt{30}$, $\sqrt{6}\sqrt{30}=6\sqrt{5}$, and $\sqrt{30}\sqrt{5}=5\sqrt{6}$, this process of multiplying numerator and denominator by the conjugates above will leave your numerator at each step as a linear combination of $1$, $\sqrt{5}$, $\sqrt{6}$, and $\sqrt{30}$ with integer coefficients.  So the process here isn't as bad as the general case, although still tedious to do "by hand".  The coefficients in the end will not generally be multiples of the integer denominator, but if they are then your answer will work out as in your question.  
Here it is actually quite simpler than the general version indicated above if you start trying to implement it. The first step would be multiplying top and bottom by $2\sqrt{30} + 2 + 3\sqrt{6}+3\sqrt{5}$, but then you are left only with $x+y\sqrt{30}$, so you can skip to step $3$.  This simplification will occur whenever your starting point is the less general $a\sqrt{A}+b\sqrt{B}+c\sqrt{AB}+d$.

Added: Here's another approach that will work in this $\sqrt A$, $\sqrt B$, $\sqrt{AB}$ case, which is an elaboration of the comment by dxiv.: By awareness that a process exists (as above) that will reduce your final answer to something of the form $a\sqrt{5}+b\sqrt{6}+c\sqrt{30}+d$, where $a$,$b$, $c$, and $d$ are rational numbers, you can just put that expression on the right hand side of your equation, multiply both sides of the equation by your denominator, group like terms, and match coefficients to solve for $a$, $b$, $c$, and $d$.  A solution exists (which we know from the process above), and it is unique by linear independence of $\{1,\sqrt5,\sqrt6,\sqrt{30}\}$.
