2020/009) Solve in real $(x+4)(x+5)(x+6)(x+7) = 1680$

We have as  4+7 = 5 + 6

$((x+4)(x+7))((x+5)(x+6)) = 1680$
Or $x^2+11x+28)(x^2+11x+30) = 1680$
Putting $x^2+11x + 29= t$ we get
$(t-1)(t+1) = 1680$
Or $t^2 -1 = 1680$ or $t^2 = 1681$ or $t=\pm 41$

So we get 2 equations $x^2+11x+29=41$ or $x^2+11x+29=-41$

1st equation gives $x^2+11x-12=0$ or $(x+12)(x-1) = 0$ giving $x = -12,1$

$x^2+11x+29=-41$ giving $x^2+11x+70=0$ does not have real root