Since squaring both sides introduced false solutions, we have to check the four solutions above by inserting them into the original equation, which shows only two of them a true solutions:
$$
\frac{1-\sqrt{21}}{2}
$$
and
$$
\frac{-1+\sqrt{17}}{2}
$$

Oh, this is a better strategy: https://www.mathportal.org/formulas/algebra/solalgebric.php

With $$a_1=0, a_2=-10, a_3=-1, a_4=20$$ the cubic equation becomes

$$
y^3+10y^2-80y-801=0
$$

and with $$y_1 = -9$$ ([rational root test](https://www.mathportal.org/calculators/solving-equations/polynomial-equation-solver.php?val1=y%5E3%2B10y%5E2-80y-801&val2=0&val3=%7By%7D%5E%7B3%7D%2B10%7By%7D%5E%7B2%7D-80y-801%3D0)) the quadratic equations become

$$
z^2+z-5=0
$$
$$
z^2+z-4=0
$$
$$
z^2-z-5=0
$$
$$
z^2-z-4=0
$$

The solutions are

$$
\frac{\pm 1 \pm \sqrt{21}}{2}
$$

and

$$
\frac{\pm 1 \pm \sqrt{17}}{2}
$$


Scroogey

science

Brain Not Braining: Part 4

noknees

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...