The question is $x^3 - 20x^2 = 8x^2 - 180$.
What I've tried is moving everything to one side, so, $x^3 - 28x^2 + 180 = 0$. Then, solving this like I would solve a cubic, I tried to find all the rational roots. After using the Rational Root Theorem, I found none.
I tried using Mathway to solve but they just told me to graph it and then find the solutions, which were really long decimals.
Is there any way of like, special factoring or a trick to solve this by hand?
It is true that the solutions of cubic equations can be ugly (have a look here) : however, this is true only if you use Cardano method.
Follow the steps given here : you have $$\Delta=14930640 \qquad\qquad p=-\frac{784}{3}\qquad\qquad q=-\frac{39044}{27}$$ So, three real roots. So, use the trigonometric solution to get $$x_k=\frac{28}{3} \left(1+2 \cos \left(\frac{1}{3} \left(2 \pi k-\cos ^{-1}\left(\frac{9761}{10976}\right)\right)\right)\right)$$ with $k=0,1,2$.
Is it so ugly ?