14. For the quadratic equation in $x$, $x^{2}+z_{1} x+z_{2}+m=0$, where $z_{1}, z_{2}, m$ are all complex numbers, and $z_{1}^{2}-4 z_{2}=$ $16+20 \mathrm{i}$. Let the two roots of this equation be $\alpha, \beta$, and $\alpha, \beta$ satisfy $|\alpha-\beta|=2 \sqrt{7}$. Find the maximum and minimum values of $|m|$.
14. By Vieta's formulas, we get $\alpha+\beta=-z_{1}, \alpha \beta=z_{2}+m$. Since $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=z_{1}^{2}-4 z_{2}-4 m$. Therefore, $|\alpha-\beta|^{2}=\left|4 m-\left(z_{1}^{2}-4 z_{2}\right)\right|=28$, which means $|m-(4+5 i)|=7$. From this, it is easy to find that $|m|_{\max }=\sqrt{41}+7 ;|m|_{\text {min }}$ $=7-\sqrt{41}$.