If $f(5+x)=f(5-x)$ for every real $x$, and $f\left(x\right)=0$ has four distinct real roots, then the sum of these roots is

The given functional equation is:

$f(5+x)=f(5-x)$

This indicates that the function $f\left(x\right)$ is symmetric about the vertical line $x=5$.

If $r$ is a root of $f\left(x\right)=0$, then we can write $r=5+d$ for some real number $d$. Since the function is symmetric about $x=5$, the value $5-d$ must also be a root of the equation.

Thus, the roots occur in symmetric pairs about $5$. Since there are four distinct real roots, we can write them as two pairs:

$x_1=5+d_1,x_2=5-d_1$

$x_3=5+d_2,x_4=5-d_2$

The sum of these four roots is:

$x_1+x_2+x_3+x_4=(5+d_1)+(5-d_1)+(5+d_2)+(5-d_2)$

$\text{Sum}=5+5+5+5=20$