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 $8072 = 4 \cdot 2018$.

I get $$ \left(x^2 + n^2 \right)^2 - \left(2nx+1 \right)^2 = (x^2 - n^2)^2 - 4nx-1 $$
Once written as a difference of squares, note that $A^2 - B^2 = 0$  means $(A+B)(A-B) = 0,$ so that  $A+B=0$  or $A-B = 0,$  possibly both.
Here  $$ n = 2018 $$
