Almost even function!

Let $f(x) = (x-a)(x-b)(x-c)(x-d)$ with $a,b,c,d \in \mathbb{R}.$ If $f(10+x) = f(10-x)$ , then:

$[(10-a)+x][(10-b)+x][(10-c)+x][(10-d)+x] = [(10-a)-x][(10-b)-x][(10-c)-x][(10-d)-x]$ (i).

By Vieta's Formulae, the sum of the four zeros of $f$ is computed via:

$(10-a) + (10-b) + (10-c) + (10-d) = -(10-a) - (10-b) - (10-c) - (10-d)$ ;

or $40 - (a+b+c+d) = -40 + (a+b+c+d);$

or $80 = 2(a+b+c+d);$

or $\boxed{40 = a+b+c+d}.$

@Tom Engelsman solution involves Vieta's Formulas but there is a little faster method.

Let $f\left(x\right)=Q\left(x\right)\left(x-a\right)\left(x-b\right)\left(x-c\right)\left(x-d\right):\ f\left(10-x\right)=f\left(10+x\right)$ with $a{,}b{,}c{,}d\in\mathbb{R}$

From the symmetry, we know WLOG that

$\frac{a+b}{2}=10\wedge\frac{c+d}{2}=10$

Adding these:

$\frac{a+b}{2}+\frac{c+d}{2}=10+10=20$

And finally, we get

$a+b+c+d=\fbox{40}$