$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$ Problem:
solve equation
$$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$$
 I don't look for easy solution (square booth side and things like that...) I look for some tricks for "easy" solution because:
 I would like to use substitution, but we have $3x$ and $-3x$, but I can't see it. 
Solution:

$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$
Or $$(\sqrt{2x^2 + 3x +5}-7) + (\sqrt{2x^2-3x+5}-5)=3x-12$$
Or $$\frac{\left(x-4\right)\left(2x+11\right)}{\sqrt{2x^2+3x+5}+7}+\frac{\left(x-4\right)\left(2x+5\right)}{\sqrt{2x^2-3x+5}+5}=3(x-4)$$
It is not difficult to prove $$\frac{2x+11}{\sqrt{2x^2+3x+5}+7}+\frac{2x+5}{\sqrt{2x^2-3x+5}+5}>3$$
Then you have $x=4$ is a root.

Another way:
Let $(\sqrt{2x^2 + 3x +5} ; \sqrt{2x^2-3x+5})\rightarrow (a;b)$ where $a;b\ge0$
So $$\frac{a^2-b^2}{2}=\frac{2x^2+3x+5-2x^2+3x-5}{2}=3x$$
Then you have a new equation $$a+b=\frac{a^2-b^2}{2}$$
Or $$-\frac{1}{2}(a-b-2)(a+b)=0$$
Can you solve it ?
