Solve 6th degree polynomial: $(x^2 - 3x - 4)(x^2 - 5x + 6)(x^2 + 2x) + 30$

Let $P(x)$ denote the LHS. Note that the first summand factors nicely: $$ P(x)=(x-2)(x-3)(x-4)x(x+1)(x+2)+30. $$ From this it is clear that $P(x)$ is symmetric around $x=1$. Explicitly, setting $y=x-1$ we have $$\begin{eqnarray*} P(x)&=&(y-1)(y-2)(y-3)(y+1)(y+2)(y+3)+30\\ &=&(y^2-1)(y^2-4)(y^2-9)+30. \end{eqnarray*}$$ Note that this is a cubic in $y^2$, so it is possible to solve for $y^2$ using the formula for roots of a cubic, and hence find the roots of $P$ in terms of radicals.

Your $6$'th degree polynomial does not have any rational roots (which you could show using the Rational Roots Theorem). It turns out to have a quadratic factor $x^2-2x-5$. It factors completely into factors of the form $x - (a + b \sqrt{6} + c \sqrt{10})$ with $a$, $b$, $c$ rational, but I don't see how you could guess that "by hand".