Solve for $x$: $6(x-3)^3 + 4(x+9)^2 +8x -5x^2 (x-10) = 0$ Solve for $x$ in:
$$6(x-3)^3 + 4(x+9)^2 +8x -5x^2 (x-10) = 0$$
So far I've made it to: $x^3 +242x +162 = 0,$  but now I'm stuck.
 A: You could use Cardano method for solving the cubic for sure ... but it will not be very pleasant; just to give you a taste, the only real solution is given by $$x=\frac{\sqrt[3]{\sqrt{43048905}-729}}{3^{2/3}}-\frac{242}{\sqrt[3]{3
   \left(\sqrt{43048905}-729\right)}}$$
If, instead, you consider the function $$f(x)=x^3 +242x +162 $$ what you can notice is that the first derivative is strictly positive; so the function always increases and there is only one root for $f(x)=0$. By inspection, since $f(0)=162$ and $f(-1)=-81$, you then know that the unique solution is such that $-1<x<0$. To solve it numerically, consider Newton method which is probably one of the simplest root-finders; starting from a guess $x_0$, the method will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ So, in your case, after simplification, this will give $$x_{n+1}=\frac{2 \left(x_n^3-81\right)}{3 x_n^2+242}$$ So, let us start with $x_0=-\frac 12$ (mid point in the interval). Applying the method, it will generate the following estimates : $-0.6683831102$, $-0.6681887179$, $-0.6681887176$ which is the solution for ten significant figures.
