If $(x+1)(x+3)(x+5)(x+7)= 5760$, what are the possible values of $x$?

$$ (x+1) (x+3) (x+5) (x+7) = 5760 $$

The equation has a symmetry around 4. We can use that to replace with $ y = x + 4 $, it becomes:

$$ (y-3) (y-1) (y+1) (y+3) = 5760 $$

or

$$ (y^2-9) (y^2-1) = 5760 $$

Now, replacing with $ z = (y^2-5) $, it becomes:

$$ (z-4) (z+4) = 5760 $$

or

$$ z^2-16 = 5760 $$

$$ z^2 = 5776 $$

So we find that: $z = 76$ or $ z = -76 $

and then work backwards, to find $y$ and $x$,

In general, quartics have closed form solutions, but they are not easy. For example, this image on wikipedia writes out the formulas for quartics.

In this case, if you multiply everything out, you could use the rational root theorem to factor it as @Dr.SonnhardGraubner does. This requires some arithmetic and doesn't really use the structure of the problem.

In this problem, if you look at the factors on the left, you see that the terms being added are symmetric around $4$ (in other words, $1$ and $7$, $3$ and $5$). Therefore, it makes sense to shift the variable $x$ by $4$. In other words, let $$ x=y-4. $$ After substitution, you get $$ (y-3)(y-1)(y+1)(y+3)=5760. $$ Rearranging this as differences of squares, you get $$ (y^2-9)(y^2-1)=5760. $$ Next, use the substitution $z=y^2$ to reduce the quartic to a quadratic. In other words, you get $$ (z-9)(z-1)=5760. $$ This can be solved with the quadratic formula after multiplying everything out. Alternatively, use @ypercube's approach of using symmetry again to make this even easier is a good idea too.

There is no need for symmetry, since $$(x+1)(x+7)(x+3)(x+5)= 5760$$ so $$(x^2+8x+7)(x^2+8x+15)= 5760$$ Put $t=x^2+8x+11$ and we get $$(t-4)(t+4) = 5760\implies t= \pm \sqrt{5776} = \pm 76$$ so $t^2+8t+11 = \pm 76$ so...