jmc

The quadratic $x^2-6x+5$ factors as $(x-5)(x-1)$, so it crosses the $x$-axis at $1$ and $5$. Since the leading coefficient is positive, it opens upwards, and thus the value of the quadratic is negative for $x$ between $1$ and $5$. Thus if $x\le 1$ or $x\ge 5$, we have $|x^2-6x+5|=x^2-6x+5$. We can solve the system in this range by setting the $y$-values equal, so

$$\begin{array}{rl}{x}^2-6x+5& =\frac{29}{4}-x\\ x^2-5x+\frac{20}{4}-\frac{29}{4}& =0\\ x^2-5x-\frac{9}{4}& =0.\end{array}$$Thus by the quadratic formula, $$x=\frac{-(-5)\pm \sqrt{(-5{)}^2-4(\frac{-9}{4})(1)}}{2(1)}=\frac{5\pm \sqrt{25+9}}{2}=\frac{5\pm \sqrt{34}}{2}.$$A quick check shows that both solutions have either $x<1$ or $x>5$, so they are both valid in this system. We do not need to find the corresponding $y$-values since the problem asks only for the sum of the $x$-coordinates.

If $1\le x\le 5$, we know $|x^2-6x+5|=-x^2+6x-5$. Solving the system as before, we have

$$\begin{array}{rl}\frac{29}{4}-x& =-x^2+6x-5\\ x^2-7x+\frac{29}{4}+\frac{20}{4}& =0\\ x^2-7x+\frac{49}{4}& =0\\ (x-\frac{7}{2}{)}^2& =0\\ x& =\frac{7}{2}.\end{array}$$Checking, this value is indeed between $1$ and $5$, so it is allowable. Thus the possible $x$-values are $\frac{5+\sqrt{34}}{2}$, $\frac{5-\sqrt{34}}{2}$, and $\frac{7}{2}$. Their sum is $$\frac{5+\sqrt{34}}{2}+\frac{5-\sqrt{34}}{2}+\frac{7}{2}=\frac{5+5+7}{2}=\boxed{\frac{17}{2}}.$$