<s>[INST] $\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$ 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:
 [/INST] 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 $3x\ge 0 \Rightarrow x\ge 0$.
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