$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$ 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:
 A: Render
$(\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5})(\sqrt{2x^2 + 3x +5} - \sqrt{2x^2-3x+5})=6x$
from the difference of squares factorization.  So then
$\color{blue}{\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x}$
$\color{blue}{\sqrt{2x^2 + 3x +5} - \sqrt{2x^2-3x+5}=2}$
Add them up:
$2\sqrt{2x^2 + 3x +5} = 3x+2$
Then square and continue.
Be sure to check the roots of your eventual quadratic equation for the proper signs of the square roots.  Can you tell from the blue equations that the proper signs will require $x>(2/3)$?
A: Let $2x^2-3x+5=t^2$. Then:
$$\sqrt{t^2+6x}+t=3x \Rightarrow t^2+6x=t^2-6xt+9x^2 \Rightarrow \\
2t=3x-2 \Rightarrow 4t^2=9x^2-12x+4 \Rightarrow \\
4(2x^2-3x+5)=9x^2-12x+4 \Rightarrow \\
x^2=16 \Rightarrow x=4.$$
Note: The square roots of the LHS of the original equation exist for all $x\in \mathbb R$, however, the LHS is nonnegative, hence $3x\ge 0 \Rightarrow x\ge 0$.
A: $\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$
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 ?
