Let $a,b,c $ be the lengths of the three sides of a triangle and $a,b$ be  the two roots of the equation $ax^2-bx+c=0 $$ (a<b) . $    Find the value range of  $ a+b-c .$
Given the problem, we need to find the value range of \(a + b - c\) where \(a, b, c\) are the lengths of the sides of a triangle and \(a, b\) are the roots of the quadratic equation \(ax^2 - bx + c = 0\) with \(a < b\).

1. **Roots of the Quadratic Equation:**
   Since \(a\) and \(b\) are the roots of the quadratic equation \(ax^2 - bx + c = 0\), by Vieta's formulas, we have:
   \[
   a + b = \frac{b}{a} \quad \text{and} \quad ab = \frac{c}{a}
   \]
   This simplifies to:
   \[
   a + b = \frac{b}{a} \quad \text{and} \quad ab = \frac{c}{a}
   \]

2. **Expressing \(b\) and \(c\) in terms of \(a\):**
   From \(a + b = \frac{b}{a}\), we get:
   \[
   b = \frac{a^2}{1 - a}
   \]
   From \(ab = \frac{c}{a}\), we get:
   \[
   c = a^2b = a^2 \cdot \frac{a^2}{1 - a} = \frac{a^4}{1 - a}
   \]

3. **Range of \(a\):**
   Since \(a < b\), we have:
   \[
   a < \frac{a^2}{1 - a}
   \]
   Solving this inequality:
   \[
   a(1 - a) < a^2 \implies a - a^2 < a^2 \implies a < 2a^2 \implies a < \frac{1}{2}
   \]
   Also, since \(a, b, c\) are sides of a triangle, they must satisfy the triangle inequalities:
   \[
   a + b > c, \quad a + c > b, \quad b + c > a
   \]
   Substituting \(b\) and \(c\) in terms of \(a\):
   \[
   a + \frac{a^2}{1 - a} > \frac{a^4}{1 - a}
   \]
   Simplifying:
   \[
   a(1 - a) + a^2 > a^4 \implies a - a^2 + a^2 > a^4 \implies a > a^4
   \]
   This implies:
   \[
   a < \frac{\sqrt{5} - 1}{2}
   \]

4. **Finding \(a + b - c\):**
   We have:
   \[
   a + b - c = a + \frac{a^2}{1 - a} - \frac{a^4}{1 - a}
   \]
   Simplifying:
   \[
   a + b - c = a + \frac{a^2 - a^4}{1 - a} = a + a^2(1 + a)
   \]
   Let \(P(a) = a + a^2(1 + a) = a + a^2 + a^3\).

5. **Range of \(P(a)\):**
   Since \(a\) is in the range \(\frac{1}{2} < a < \frac{\sqrt{5} - 1}{2}\), we need to evaluate \(P(a)\) at these endpoints:
   \[
   P\left(\frac{1}{2}\right) = \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{4 + 2 + 1}{8} = \frac{7}{8}
   \]
   \[
   P\left(\frac{\sqrt{5} - 1}{2}\right) = \frac{\sqrt{5} - 1}{2} + \left(\frac{\sqrt{5} - 1}{2}\right)^2 + \left(\frac{\sqrt{5} - 1}{2}\right)^3
   \]
   Since \(P(a)\) is strictly increasing for \(a > 0\), we have:
   \[
   P\left(\frac{1}{2}\right) < P(a) < P\left(\frac{\sqrt{5} - 1}{2}\right)
   \]
   Therefore:
   \[
   a + b - c \in \left(\frac{7}{8}, \sqrt{5} - 1\right)
   \]

The final answer is \( \boxed{ \left(\frac{7}{8}, \sqrt{5} - 1\right) } \)