Solving $(x+1)\sqrt{2(x^2 + 1)} + \sqrt{6x^2 + 18x +12} = \frac{3x^2 + 7x + 10}{2}$ Today, I came across this problem.
$$(x+1)\sqrt{2(x^2 + 1)} + \sqrt{6x^2 + 18x  +12} = \dfrac{3x^2 + 7x + 10}{2}$$
We are asked to find the possible values of $x$ satisfying this equation.
The first thought which came to my mind is to use some kind of factorisation. I tried for like an hour but all in vain.
Then, I tried to solve the problem by multiplying both sides by 2 and then squaring both sides. The equation became too complicated.
By using some hit and trial, I get to know that $x = 1$ satisfies the equation. But what about complex solutions. So this method is also of no use.
I am sure this question has to be solved using some special equality which I'm unaware of. I want a method so that, I could obtain all the possible values. Can anyone help me or just give some hints?

Square the equation rearrange & square again. This will allow you to remove the radicals & leave you with a polynomial.
\begin{eqnarray*}
(3x^2+7x+10)^2-8(x+1)^2(x^2+1)-4(6x^2+18x+12)=8(x+1)\sqrt{2(x^2+1)(6x^2+18x+12)} \\
\end{eqnarray*}
\begin{eqnarray*}
 ((3x^2+7x+10)^2-8(x+1)^2(x^2+1)-4(6x^2+18x+12))^2-128(x+1)^2(x^2+1)(6x^2+18x+12)=0.
\end{eqnarray*}
Now I would resort to CAS ...

So $x=1$ is the only real solution as $x^6+54x^5+153x^4+104x^3 -72x^2+400$ is always positive. The complex roots will be jolly unpleasant & unilluming.
