TITLE: If $(x+1)(x+3)(x+5)(x+7)= 5760$, what are the possible values of $x$?
QUESTION [6 upvotes]: I was doing some questions related to quadratic equations and I stumbled upon this question: Solving the equation
$$(x+1)(x+3)(x+5)(x+7)= 5760.$$
One of the hints was to convert it into a quadratic equation.
I tried to but couldn't get anything meaningful. Is there any way in which I can solve bi-quadratic equations fast?
(Sorry if this strikes as basic, I am just a highschooler.)

REPLY [3 votes]: 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...