How to solve this in an efficient way (without calculators) 
Solve for $x$, if
  $$(x+4)(x+7)(x+8)(x+11)+20=0.$$

Is there an easier way to solve this than trying to multiply all the values together? I've tried multiplying all of them together, and I get an equation of the fourth degree which I find very hard to factorise. I'm hoping there is an easier way to solve this question. Any help is appreciated.Thanks :).
The equation I get after multiplying is
$$x^4+30x^3+325x^2+1500x+2484=0,$$
and its roots are 
$$-6, \quad -9, \quad \frac{-15\pm\sqrt{41}}2 .$$
 A: $f(x) = (x+4)(x+7)(x+8)(x+11)+20\ $ is symmetric about the line $x = -7.5$
$y = x + 7.5$
$(y-3.5)(y-0.5)(y+0.5)(y+3.5)+20=0$ 
$(y^2 - 0.5^2)(y^2 -3.5^2)+20=0$ 
$y^4 - 12.5 y^2 + 23.0625 = 0$
Use the quadratic formula to solve for $y^2.$
The square roots of that give $y.$
And, subtract $7.5$ to get $x.$
A: Yes, it can be solved more easily.

(1) Let $u = x + {\small{\displaystyle{\frac{15}{2}}}}$.

(2) Replace $x$ by $u - {\small{\displaystyle{\frac{15}{2}}}}$.

(3) Group the $4$ factors in matching pairs.

(4) Expand the matched pairs.

(5) Solve for $u^2$.

(6) Solve for $u$.

(7) For each solution, set $x = u - {\small{\displaystyle{\frac{15}{2}}}}$.
A: If we're looking for integer solutions, we can see that $x + 4$, $x + 7$, $x + 8$, and $x + 11$ must be four integers whose product is $-20$, and so each must be $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$. The difference between the smallest and largest factor is $(x + 11) - (x + 4) = 7$, so those two factors must be either $-5$ and $2$ or $-2$ and $5$. These values correspond to the possibilities $x = -9, -6$, and substituting gives that these are indeed solutions.
With these in hand, one can expand the quartic into standard form and divide by the linear factors $x + 6$ and $x + 9$ we now know it has, leaving a quadratic to which we may apply the quadratic formula to find the remaining roots.
Of course, most quartic polynomials are irreducible over $\Bbb Q$, and so most quartic equations cannot be solved this way.
A: Rearranging the LHS, 
$$(x+4)(x+11)(x+7)(x+8)+20=0$$
to have
$$(x^2+15x+44)(x^2+15x+56)+20=0$$
Now setting $x^2+15x=t$ gives
$$(t+44)(t+56)+20=0,$$
i.e.
$$t^2+100t+2484=0\iff (t+54)(t+46)=0$$
Now solve
$$x^2+15x=-54,-46$$
