Find this following Diophantine equation all integer solution $$(3x-1)^2+2=(2y^2-4y)^2+y(2y-1)^2-6y$$ or $$9x^2-6x+3=4y^4-12y^3+12y^2-5y$$ Maybe this equation can be solved by using Pell equation methods?
I want take right is Quadratic formula with $(ay^2+by+c)$
maybe we have $$(3x-1)^2-A(ay^2+by+c)^2=B?$$
You can write it as
$$(3x-1)^2-(2y^2-3y+\tfrac34)^2=-\tfrac12y-\tfrac{41}{16}.$$
Factoring the LHS gives two factors at least one of which gets too large as $y$ is large, as
$$|(3x-1)-(2y^2-3y+\tfrac34)|+|(3x-1)+(2y^2-3y+\tfrac34)|\geqslant2\cdot|2y^2-3y+\tfrac34|.$$
It suffices to check the $y$'s with $-\tfrac12y-\tfrac{41}{16}=0$ (impossible) or $|-\tfrac12y-\tfrac{41}{16}|\geqslant|2y^2-3y+\tfrac34|$, that is, $y\in\{0,1,2\}$.
(Note all this makes a little more sense after denominators are cleared, but it's perfectly valid to act as if they are.)
The only solution is $(1,2)$.
Yes, you can use the quadratic formula. Solve for $x$ and you get the form,
$$x = \frac{-b\pm\sqrt{f(y)}}{2a}$$
hence you must find integer $y$ such that,
$$f(y) = z^2$$
If your $f(y)$ is a polynomial only of degree $n=2$, then generally you can use a Pell equation to find an infinite more integer $y$.
Unfortunately, your $f(y)$ has degree $n>2$ (namely $n=4$), and after some transformations this is an elliptic curve. And it is well-known an elliptic curve only has a finite number of integer solutions.
