How to resolve this simple pde?

Question

I am given the following PDE: $$u_{xx} + u_{xt} − 20u_{tt} = 0.$$ On factorizing the operators we get that $$(∂_x + 5∂_t)(∂_x − 4∂_t)u = 0.$$ If we let $$\alpha = ax+bt$$ $$\beta = cx+dt.$$ Then by the chain rule $$\partial_{x} = a\partial_{\alpha}+b\partial_{\beta}$$ and $$\partial_{t} = c\partial_{\alpha}+d\partial_{\beta}.$$ Thus $$(∂_x + 5∂_t)(∂_x − 4∂_t) = ((a+5c)\partial_{\alpha+}+(b+5d)\partial_{t})((a-4c)\partial_{\alpha+}+(b-4d)\partial_{t}).$$ We choose $c=1,a=4,d=1$ and $b=-5.$ Then
$$(∂_x + 5∂_t)(∂_x − 4∂_t) = (9\partial_{\alpha})(-9\partial_{\beta})=0.$$ Thus $$u(x,y)=f(\alpha)+g(\beta)=f(4x-5t)+g(x+t).$$ We check this $$u_{xx} = 16f''+g$$ $$u_{xt} = -20f''+g''$$ $$u_{tt} = 25f''+g''.$$ On substituting this solution does not work. Where have I made a mistake?