Solve $(x^2-2018^2)^2 - 8072x - 1 = 0$ I already know the answer using WolframAlpha, however, I don't know how to tackle this by hand.
I've tried using the difference of squares on the first term, didn't get much though, I tried using $x^2-2018^2 = \sqrt{8072x+1}$, but I didn't know how to continue without squaring again, and I've noticed $8072 = 4 \cdot 2018$.

Another approach:
$$(x^2-2018^2)^2=8072 x+1$$
$$(x-2018)^2(x+2018)^2=4\times 2018 x+1$$
A:
$ \begin{cases}(x-2018)^2=4\times 2018 x+1 \\ (x+2018)^2=1\end {cases}$
$x=-2019$ and $x=-2017$ which are the solution of $(x+2018)^2=1$  can be solutions of this system if they suffice the first equation. This gives a family of solutions.
B:
$ \begin{cases}(x+2018)^2=4\times 2018 x+1 \\ (x-2018)^2=1\end {cases}$
Similarly this system can  gives another family of solutions.
