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well that formula they showed is actually derived from the original quartic formula after depressing it and going case wise depending on the monic polynomial which is not the case for this one, this one is a quasi-symmetric variant, so it's not likely to give the answer :)
Good try tho! Most people won't reach till here...
Oh, this is a better strategy: https://www.mathportal.org/formulas/algebra/solalgebric.php
With the cubic equation becomes
and with (rational root test) the quadratic equations become
The solutions are
and
How did you determine the coefficients so arbitrarily? The answer is correct though but I think that's cheating. Using factor theorem is a very bad way to solve this kind of equation. Anyways 200 sats for the good effort and somehow getting the correct answer :)
that's okay, but the rational root test link shows you solved the equation using a solver which is not okay, had you not posted the link and only written "rational root test", I might not have figured it out, but since you did, it puts a wrong impression
it's okay to use a solver once you understand the method, but citing is not always necessary! (another lesson I learnt the hard way)
but another 200 sats for being honest and using formulas the correct way!
I would have squared both sides to
and then followed https://math.stackexchange.com/questions/785/general-formula-for-solving-quartic-degree-4-equations/1219804#1219804 with a=0, b=-10, c=-1, d=20. But the quadratic equation
has no solution, indicating that I made a mistake, and I grew tired :-)