China Mathematical Competition

$x^2 - a x - 4 = 0$ has two roots:
$$x_1 = \frac{a}{2} - \sqrt{4 + \frac{a^2}{4}}, \quad x_2 = \frac{a}{2} + \sqrt{4 + \frac{a^2}{4}}$$
We have $B \subseteq A \Leftrightarrow x_1 \geq -2$ and $x_2 < 4$. This means that
$$\frac{a}{2} - \sqrt{4 + \frac{a^2}{4}} \geq -2, \quad \frac{a}{2} + \sqrt{4 + \frac{a^2}{4}} < 4$$
From the above we get $0 \leq a < 3$.