How to solve $x(3x+3)(x+5)(2x+12)+576 = 0$?

It is equivalent to $x(x+1)(x+5)(x+6)+96 = 0$

Now $$(x^2+6x)(x^2+6x+5)+96=0$$

Let $t=x^2+6x$ and finish the job...

As already said, $$x(3x+3)(x+5)(2x+12)+576 = 0$$ is equivalent to $$x(x+1)(x+5)(x+6)+96 = 0.$$ Note the symmetry of the set of numbers $0,1,5,6.$
Set $a=x+3,$ the equation is equivalent to $$\begin{aligned}(a-3)(a-2)(a+2)(a+3)+96=&0\\ (a^2-4)(a^2-9)+96=&0\\ a^4-13a^2+132=&0 \end{aligned}$$ With the methods you know it is easy to finish.

Let $f(x)=x(3x+3)(x+5)(2x+12)$. Then $f'(x)=24x^3+216x^2+492x+180$, whose roots are $-3$ and $\frac12\left(-6\pm\sqrt{26}\right)$. But $f(-3)=216$ and $f\left(\frac12\left(-6\pm\sqrt{26}\right)\right)=-\frac{75}2$. Therefore, the absolute minimum of $f$ is $-\frac{75}2$ and so the absolute minimum of $x(3x+3)(x+5)(2x+12)+576$ is greater than $0$.