Question: find the general solution of $u_{xx} + 2u_{xt} - 20u_{tt} = 0$
I'm bit confused how exactly the form of general solution $u(x,t)$ looks like. Because what I learn from the book is when wave equation is $u_{tt} - c^2u_{xx} = 0$. Then $u(x,t) = f(x - ct) + g(x - ct)$ where $f$ and $g$ are arbitrary function and $c$'s are two roots. However, I'm referring to few websites here and the equation $u_{xx} +u_{xt} - 20u_{tt} = 0$. This,https://www.math.cuhk.edu.hk/course_builder/1516/math4220/Solutions%20to%20assignment%202.pdf by factoring the equation into $(\frac{\partial }{\partial x} - 4\frac{\partial }{\partial t})(\frac{\partial }{\partial x} + 5 \frac{\partial }{\partial t}) = 0$ and obtained the general solution $u(x,t) = f(x + \frac{1}{4}t) + g(x - \frac{1}{5}t)$. I don't understand why $c$ became $\frac{1}{c}$ here.
In addition, this link http://individual.utoronto.ca/rifkind/teaching/PDE2015/Assignments/Assignment2/solution2.pdf using changing coordinate method to find the general solution $u(x,t) = f(t - 5x) + g(4x + t)$. These two solutions look equivalent to me, though I can't tell why at this moment.
Now, back to my question, it's bit trickier than the one I'm referring to. Since it can't factor in general way, but we can still 'factor' using quadratic formula to find the roots and expressed into
$(D_{x}^2 + 2D_{xt} - 20D_{t}^2)u = 0$
$(D_x - (-1 + \sqrt{21})D_t)(D_x - (- 1 - \sqrt{21})D_t) = 0$
$(D_x + (1 - \sqrt{21})D_t)(D_x + ( 1 + \sqrt{21})D_t) = 0$
I think I'm stuck on the very last part which is writing down the general solution $u(x,t)$.
Any suggestion?