Number theory: natural sqrt of 2nd degree polynomial above the naturals.

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I am trying to find solutions to $y =\sqrt{ax^2+bx+c}, x \land y \in \mathbb{N}$.

I've looked at Sqrt of polynomial, How to find integer X that gives integer y

but the solution there assumes $a=1$, and fails when $\sqrt{a} \notin \mathbb{N}$

because $(y+\sqrt{a}x+\frac{b^2}{a})(y-\sqrt{a}x-\frac{b^2}{a})=\frac{b^2}{a}+c$ doesn't imply that the factors in LHS equal factors of RHS.

assuming i want the first $k$ solutions, methods of reducing the number numbers to check are also welcome!

I'm thinking that maybe there is some modulus you can take, that will simplify a lot. maybe with some Legendre symbol analysis thrown in. but maybe its just because i learnt these things recently XD

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solved by individ in the comments!

using pells equation to solve: https://artofproblemsolving.com/community/c3046h1048216__

solving pells equation: https://www.youtube.com/watch?v=_xGWbZte1Ds