<s>[INST] How to solve the equation $\sqrt{x^{2}-x+2}+\sqrt{2x^{2}+2x+3}=\sqrt{2x^{2}-1}+\sqrt{x^{2}-3x-2}$ 
Solve in real numbers the equation $$\sqrt{x^{2}-x+2}+\sqrt{2x^{2}+2x+3}=\sqrt{2x^{2}-1}+\sqrt{x^{2}-3x-2}.$$

 [/INST] Rewrite the equation as 
$$\sqrt{x^{2}-x+2}-\sqrt{x^{2}-3x-2}=\sqrt{2x^{2}-1}-\sqrt{2x^{2}+2x+3}.\tag{$1$}$$
On each side, do the usual trick of multiplying "tops" and "bottoms" by the conjugates. A semi-miracle happens. We get
$$\frac{2x+4}{ \sqrt{x^{2}-x+2}+\sqrt{x^{2}-3x-2} }=\frac{-(2x+4)}{\sqrt{2x^{2}-1}+\sqrt{2x^{2}+2x+3}}.\tag{$2$}$$
After making  note to check about $x=-2$ later, cancel the $2x+4$ terms on top of $(2)$, then flip both sides of $(2)$ over. 
We get an equation that marries very nicely with $(1)$, and two of the square roots are gone.  We get an equation of shape $\sqrt{A}=-\sqrt{B}$. So it comes down to checking $x=-2$ (works).  
Remark: The trick could have been used directly with the equation as given. However, in more complicated cases, reduction of degree is useful.  
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