I"m supposed to solve this equation. It"s from a hyakkendana-hashigozake.com contest so solving it by hand would be preferable (no quartic formulas).I thought about making $u = x^2-3x-2$ obviously but it leads to another quartic equation. I also tried the substitution $u=x+2$, and after the whole expand trinomial, simplify, invoke rational root theorem and test roots, I still got nothing out of it.

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I noticed that $x^2-3x-2$ can"t be factored nicely so I dunno what other route to take. Lots of equations I tackled in hyakkendana-hashigozake.com contests could make use of nice trigonometric substitutions, but none in particular pop in my head right now.

If anyone can give me hints or a full solution, that would be awesome. Thanks!

Since you have a quartic, there are potentially 4 real solutions.

Note that any solution to $x = x^2 - 3x - 2$ is also a solution to the given equation.

Hence, $x^2 - 4x - 2$ is a factor of the quartic. Now find the other factor, and solve both quadratics.

Alternatively, see this almost 10 year old thread on the Art of Problem Solving Forum, or this slightly newer thread which discusses a similar problem.

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