<s>[INST] How many different solutions for this $\sqrt{3x^2 + 2x + 5} + \sqrt{x^2-4x+5} = \sqrt{2x^2 - 2x + 2} + \sqrt{2x^2+8}$
My attempt :
All the equation inside sqrt has $D<0$.
Let it be
$\sqrt A + \sqrt B = \sqrt C + \sqrt D$
2 possibilities
$$\sqrt A = \sqrt C \cap \sqrt B = \sqrt D$$
Find $x$ that match this condition. 
$\sqrt A = \sqrt C $ i get $x=-3 \cup x=-5$
$\sqrt B = \sqrt D $ i get $x=-3 \cup x=-1$
$$\sqrt A = \sqrt D \cap \sqrt B = \sqrt C$$
Find $x$ that match this condition.
$\sqrt A = \sqrt D $ i get $x=-3 \cup x=1$
$\sqrt B = \sqrt C $ i get $x=-3 \cup x=1$
2 solutions.
Another posibility
$\sqrt A + \sqrt B = \sqrt C + \sqrt D$
${[\sqrt A + \sqrt B]}^2 = {[\sqrt C + \sqrt D]}^2$
This part is complicated.
Is there easier way to solve it?
 [/INST] Let $a=\sqrt{3x^2+2x+5}, b=\sqrt{2x^2+8}, c=\sqrt{2x^2-2x+2}, d=\sqrt{x^2-4x+5}$
$\implies a-b=c-d$ 
Observe that $$3x^2+2x+5-(2x^2+8)=x^2+2x-3=2x^2-2x+2-(x^2-4x+5)$$
$\implies a^2-b^2=c^2-d^2$
$\implies a+b=\dfrac{c-d}{a-b}\cdot c+d=c+d$
Now add and subtract to find $$a=c$$ 
$$\implies\sqrt{3x^2+2x+5}=\sqrt{2x^2-2x+2}$$
Square both sides
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