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

This was a question on a test i recently took, And i wasn't able to solve it. I later tried to solve it using online calculators and it turns out this doesn't have any real solutions.
I know there's a general formula for quartic polynomials that can work but we were only taught two methods, modifying the equation to a quadratic one using substitution or guessing some of the solutions. Both of these didn't work for me.
Are there any ways to prove that this doesn't have real solutions without the quartic formula?
 A: 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.
A: 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 $0$.
A: 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...
A: Another possible way is using the symmetry of $x(3x+3)(x+5)(2x+12) = 6x(x+1)(x+5)(x+6)$ around $x=\color{blue}{3}$:
$$6x(x+1)(x+5)(x+6) = 6(x+\color{blue}{3}-3)(x+\color{blue}{3}+3)(x+\color{blue}{3}-2)(x+\color{blue}{3}+2)$$ $$=6(\underbrace{(x+\color{blue}{3})^2}_{y:=}-9)((x+\color{blue}{3})^2-4)$$
The minimum value of $(y-9)(y-4)$ is $-\frac{25}{4}$. Hence,
$$6x(x+1)(x+5)(x+6)+576 \geq 6(-\frac{25}{4})+576 >0$$
A: Along with all other answers, you can also try the general method for any degree 4 polynomial.
If f(x) is monic and has degree 4, You can break it as:
$f(x) = (x^2 + px + q)(x^2 + rx + s)$
The idea is now to compare the coefficients.  If
$f(x) = x^4 + ax^3 + bx^2 + cx + d$
this gives the relations:
a = p + r
b = pr + q + s
c = ps + qr
d = qs
And then Check for the solution of two quadratic equations.
