$x^2 + 2ax + a^2 - 4 = 0$
$D = a^2 - (a^2 - 4) = 4 > 0$
$x_{1,2} = -a ± 2$
$\begin{equation*} \begin{cases} -6 < -a - 2 < 6 &\\ -6 < -a + 2 < 6 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} -6 + 2 < -a - 2 + 2 < 6 + 2 &\\ -6 - 2 < -a + 2 - 2 < 6 - 2 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} -4 < -a < 8 &\\ -8 < -a < 4 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} 4 > a > -8 &\\ 8 > a > -4 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} -8 < a < 4 &\\ -4 < a < 8 & \end{cases} \end{equation*}$
−4 < a < 4
Ответ: при a ∈ (−4;4)