15. Given the function $f(x)=a x^{2}+8 x+3(a<0)$, for a given negative number $a$ there is a maximum positive number $l(a)$ such that the inequality $|f(x)| \leqslant 5$ holds for all $x$ in the interval $[0, l(a)]$. Find the value of $a$ for which $l(a)$ is maximized, and determine the maximum $l(a)$.
15. $f(x)=a\left(x+\frac{4}{a}\right)^{2}+3-\frac{16}{a}$, so $f(x)_{\max }=3-\frac{16}{a}$.
(1) $3-\frac{16}{a}>5$ which is $-8-\frac{4}{a}$, so $l(a)$ is the larger root of $a x^{2}+8 x+3=-5$.

Thus, $l(a)=\frac{-8-\sqrt{64-32 a}}{2 a}=\frac{4}{\sqrt{4-2 a}-2} \leqslant \frac{4}{\sqrt{20}-2}=\frac{\sqrt{5}+1}{2}$.
When and only when $a=-8$, the equality holds. Since $\frac{\sqrt{5}+1}{2}>\frac{1}{2}$, therefore, when and only when $a=-8$, $l(a)$ reaches its maximum value $\frac{\sqrt{5}+1}{2}$.