10. Given $f(x)=x^{5}-10 x^{3}+a x^{2}+b x+c$, if the roots of the equation $f(x)=0$ are all real numbers, $m$ is the largest of these 5 real roots, then the maximum value of $m$ is $\qquad$
Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be the roots of the equation. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=0, \\
\sum_{1 \leqslant i<j \leqslant 5} x_{i} x_{j}=-10
\end{array} \Rightarrow x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}=20 .\right.
$$

Thus, we have $\left\{\begin{array}{l}x_{1}+x_{2}+x_{3}+x_{4}=-m, \\ x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=20-m^{2}\end{array} \Rightarrow(-m)^{2} \leqslant 4\left(20-m^{2}\right) \Rightarrow m \leqslant 4\right.$.
Equality holds when $x_{1}=x_{2}=x_{3}=x_{4}=-1, x_{5}=4$. Therefore, the maximum value of $m$ is 4.