Separate the operators and u:
\begin{equation} (A\partial_{t}^{2} + 2B\partial_{t}\partial_{x} + C\partial_{x}^{2})u = 0 \end{equation}
Provided \begin{equation} 4B^{2} - 4AC >= 0 \end{equation} we have roots r1, r2 s.t. \begin{equation} \partial_{t} = r_1\partial_x, \partial_t = r_2\partial_x\end{equation}
if equal, then r1 = r2.
then we can transform the original equation to form: \begin{equation}(\partial_{t} - r_1\partial_x)(\partial_t - r_2\partial_x)u = 0\end{equation}
we can solve this equation in the way taught in class, namely substitute the partial part and u to v and w then get u. The answer is: \begin{equation} u = f(x+r_1t) + g(x+r_2t) \end{equation}
compare to the form given by the question, we have \begin{equation}r_1 = -c_1, r_2 = -c_2 \end{equation} are the roots.
a). so they satisfy \begin{equation} c_1 + c_2 = \frac{2B}{A}, c_1c_2 = \frac{C}{A}\end{equation}.
b). to have \begin{equation} c_1 < c_2\end{equation} we need \begin{equation} 4B^2 - 4AC = B^2 - AC > 0\end{equation}
