Solve the equation by putting new variable I want to solve the equation
$$x^2+(2 x+3) \sqrt{3 x^2+6 x+2}=6 x+5.$$
I tried. Put $t =  \sqrt{3 x^2+6 x+2}.$ We have
$$x^2+(2 x+3)t - 6x - 5 = 0.$$
Discriminant of this equation (with unknown $x$) is $56-36t+4t^2$. It is not a square number. 

Just put it like this:
$(2 x + 3) \sqrt{(3 x^2 + 6 x + 2)} = (6 x + 5 - x^2)$
then square both sides:
$(2 x + 3)^2 (3 x^2 + 6 x + 2) = (6 x + 5 - x^2)^2$
and collect the terms:
$(-1 + x (5 + x)) (7 + x (17 + 11 x))=0$
The answer is straightforward. Will this help?:)
P.S.: the solutions are 
$\left\{x\to \frac{1}{22} \left(-17-i \sqrt{19}\right)\right\},\left\{x\to \frac{1}{22} \left(-17+i \sqrt{19}\right)\right\},\left\{x\to \frac{1}{2} \left(-5-\sqrt{29}\right)\right\},\left\{x\to \frac{1}{2} \left(\sqrt{29}-5\right)\right\}$
