Looking for a trick to solve $2\sqrt {2x}+\sqrt {2x+3}=\sqrt {3x+2}+\sqrt {6x+20}$ Consider the equation:
$$2\sqrt {2x}+\sqrt {2x+3}=\sqrt {3x+2}+\sqrt {6x+20}.$$
Find a trick ( if exists ) which allows to solve it elegantly i.e. with avoiding  the  systematic squaring.
(The systematic squaring inevitably leads to a fourth-degree equation:
$$
\begin{align}
0
&=
207x^4-12564x^3+27738x^2+231084x-40401\\[6pt]
&=9\left( 23x^2-1258x-4489\right) \left( x^2-6x+1\right)\;,
\end{align}
$$
so the answer is 
$$x=\dfrac {629+\sqrt {498888}} {23}.$$

*

*This equasion is defined for all $x>0$.

*Function on the left hand side is strictly increasing (i.e always increasing).

*Function on the right hand side is strictly increasing (i.e always increasing) too.
We can draw a simple graph on large scale, but with low prescision, to find out where too look for this root. We'll see on the graph that there is 1 intersection. Beacuse both functions strictly increasing there will be no other roots for $x>0$.
You can use this or any other tool to draw two graphs for left and right hand side functions.
From that graph we'll see that $x \in (58.02 , 58.08)$. You can get the decimal answer depending on the prescision required by incrasing prescision of interval ends. 
