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
$$
\frac{-1+\sqrt{17}}{2}
$$

Scroogey

science

Brain Not Braining: Part 4

noknees

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

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