14. (9th "Hope Cup" Invitational Training Question) Given the function $f(x)=a x^{2}+b x+c(a \neq 0)$ defined on the interval $\left[-\frac{b}{2 a}, \frac{2 a-b}{2 a}\right]$, and the minimum value of this function is not $\frac{4 a c-b^{2}}{4 a}$, then the minimum value of this function is $\qquad$ .
14. Fill $\frac{4 a^{2}+4 a c-b^{2}}{4 a}$. Reason: On the interval $\left[-\frac{b}{2 a}, \frac{2 a-b}{2 a}\right]$, the function $y=a x^{2}+b x+c$ is monotonic. Given that $\frac{4 a c-b^{2}}{4 a}$ is not the minimum value, it indicates that $a<0$, which means the function is decreasing on the interval $\left[-\frac{b}{2 a}, \frac{2 a-b}{2 a}\right]$. Therefore, the minimum value is $a\left(\frac{2 a-b}{2 a}\right)^{2}+b\left(\frac{2 a-b}{2 a}\right)+c=\frac{4 a^{2}+4 a c-b^{2}}{4 a}$