Find the sum of the roots

$(x-5)(x+4)(x-7)(x+6)=504. \\ ~~~\\ So (x-5)(x+4)(x-7)(x+6)-504=0. \\ ~~~\\ So~~x^4+(-7-5+4+6)x^3+............=0.\\ ~~~\\ -7-5+4+6=-2.\\ ~~~\\ By~ Vieta's~ formula,~ sum ~of ~roots~ is~ negative~ of~ coefficient~ of ,~x^3~term~=-(-2)=\color{#D61F06}{2}.$

The simplified equation by multiplying the terms is $\large x^4-2x^3-61x^2+62x+336=0$

By Vietas formula , sum of roots is negative of coefficient of $x^3$ , which is $-(-2)=\boxed{2}$

Rearranged and multiply.

$(x-5)(x+4)(x-7)(x+6)=504$

$(x^2-x-20)(x^2-x-42)=504$

let $a=x^2-x$ , then

$(a-20)(a-42)=504$

$a^2-62a+336=0$

By using the quadratic formula, we get $a=56$ and $a=6$

when $a=56$ ,

$56=x^2-x$ $\implies$ $x^2-x-56=0$ $\implies$ $x=8,x=-7$

when $a=6$

$6=x^2-x$ $\implies$ $x^2-x-6=0$ $\implies$ $x=3,x=-2$

The sum of the roots is $8-7+3-2=\boxed{2}$ .