Let $u, v$ be the roots of the quadratic equation $a x^{2} + b x + c = 0$ such that $0 \leq u \leq v \leq 1$. We have the relations $b = -a(u + v)$ and $c = u v$. Therefore
$$\begin{aligned} \frac{(a-b)(2a-b)}{a(a-b+c)} &= \frac{(1 + u + v)(2 + u + v)}{1 + u + v + u v} \\ &= 2 + \frac{u}{1 + v} + \frac{v}{1 + u} \\ &\leq 2 + \frac{u}{1 + u} + \frac{1}{1 + u} = 3 \end{aligned}$$
Clearly, when $u = v = 1$, the equality holds. Thus, $3$ is the maximum.