14. Let the function $f(x)=a x^{2}+8 x+3(a<0)$. For a given negative number $a$, there is a largest positive number $l(a)$, such that the inequality $|f(x)| \leqslant 5$ holds for the entire interval $[0, l(a)]$. Question: For what value of $a$ is $l(a)$ the largest? Find this largest $l(a)$. Prove your conclusion.
$$
f(x)=a x^{2}+8 x+3=a\left(x+\frac{4}{a}\right)^{2}+3-\frac{16}{a} \text {. }
$$
(1) If $l(a)5 \Rightarrow-8-\frac{4}{a}$, at this time $3-\frac{16}{a} \leqslant 5 \Rightarrow a \leqslant-8$.

To make $l(a)$ the maximum $\Rightarrow l(a)$ is the larger root of the equation $f(x)=-5$ $\Rightarrow l(a)=\frac{-8-\sqrt{64-32 a}}{2 a}$ $=\frac{\sqrt{16-8 a}+4}{-a}=\frac{4}{\sqrt{4-2 a}-2} \leqslant \frac{2}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{2}$, equality holds when $a=-8$. Therefore, when $a=-8$, the maximum value of $l(a)$ is $\frac{\sqrt{5}+1}{2}$.