Quick, expand it!

$(x-7)(x-3)(x+5)(x+1)=1680$

$(x-7)(x+5)(x-3)(x+1)=1680$

$(x^2-2x-35)(x^2-2x-3)=1680$

Let $x^2-2x=a$

$(a-35)(a-3)=1680$

$a^2-38a+105-1680=0$

$a^2-38a-1575=0$

$(a-63)(a+25)=0$

$\Rightarrow a = 63 \quad or \quad a = -25$

When $a=-25 \Rightarrow x^2-2x+25=0 \Rightarrow x=\dfrac{2\pm\sqrt{-96}}{2}$ which are complex roots .

When $a=63 \Rightarrow x^2-2x-63=0$ , the discriminant is positive.

Hence by Vieta's formula , sum of all roots $=2+2\huge=\boxed{4}$

Cheers! xD

Suppose, the solutions for the final equation are: a, b, c and d

Then, (x-a)(x-b)(x-c)(x-d) = 0

x^4 - (a+b+c+d)x^3+ .... +abcd = 0

The given equation is: (x-7)(x-3)(x+5)(x+1)-1680 = 0

The co-efficients in both the equations have to be same. Taking co-efficients of x^3:

(a+b+c+d) = (7+3-5-1) = 4

After expansion, coefficient of x^3 is -4 and that of x^4 is 1. (-b/a) gives sum of roots, which is -(-4/1) = 4

$\left( x-7 \right) \left( x-3 \right) \left( x+5 \right) \left( x+1 \right) =1680\\ \Rightarrow { x }^{ 4 }-{ 4 }x^{ 3 }-{ 34 }x^{ 2 }+76x-1575=0\\ By\quad vieta's\quad formula\quad \\ sum\quad of\quad roots=4$

Note that to find the sum of the solutions we must find $\frac{a_{n-1}}{a_{n}}$ where $a_{n-1}$ is the coefficient of $x^3$ and $a_{n}$ is the coefficient of $x^4$ . The $x^4$ coefficient is clearly 1. There are only 4 $x^3$ terms when the brackets are expanded, each formed by multiplying the $x$ terms in 3 of the brackets and then multiplying by the constant in the remaining bracket.

$(\color{#D61F06}x\color{#333333}-7)(\color{#D61F06}x\color{#333333}-3)(\color{#D61F06}x\color{#333333}+5)(x\color{#D61F06}+1\color{#333333}) \rightarrow \color{#D61F06}1x^3$

$(\color{#3D99F6}x\color{#333333}-7)(\color{#3D99F6}x\color{#333333}-3)(x\color{#3D99F6}+5\color{#333333})(\color{#3D99F6}x\color{#333333}+1) \rightarrow \color{#3D99F6}5x^3$

$(\color{#20A900}x\color{#333333}-7)(x\color{#20A900}-3\color{#333333})(\color{#20A900}x\color{#333333}+5)(\color{#20A900}x\color{#333333}+1) \rightarrow \color{#20A900}-3x^3$

$(x\color{#EC7300}-7\color{#333333})(\color{#EC7300}x\color{#333333}-7)(\color{#EC7300}x\color{#333333}-3)(\color{#EC7300}x\color{#333333}+5) \rightarrow \color{#EC7300}-7x^3$

The $x^3$ term is therefore $\color{#D61F06}1x^3 +\color{#3D99F6}5x^3 +\color{#20A900}(-3x^3) +\color{#EC7300}(-7x^3) = \color{#333333}-4x^3$

Therefore the sum of solutions is $-\frac{-4}{1} = \boxed{4}$

The first of Newton's Identities is the way to go, of course, but you can also define $x:=y+1$ to get an even polynomial $P(y)$ : $1680=(y-6)(y-2)(y+6)(y+2)=(y^2-4)(y^2-36)\quad\Rightarrow\quad P(y):=y^4-40y^2-1536=0$ By symmetry, $P(y)$ has the roots $y_{1,2}=\pm a,\:y_{3,4}=\pm b,\quad a,\:b\in\mathbb{C}$ , so they cancel each other out in the sum $\sum_{k=1}^4 x_k=\sum_{k=1}^4(1+y_k)=4+\sum_{k=1}^4 y_k =4+0=\boxed{4}$