A geometry problem by Alfa Claresta

$\tan(\alpha+\beta) = \dfrac{\tan \alpha +\tan \beta}{1-\tan \alpha \cdot \tan \beta}$

Sum of roots of $ax^2+bx+c=0$ is $-\dfrac ba$ and Product of roots is $\dfrac ca$ . As roots of the equation $x^2-4x-9=0$ are $\tan \alpha$ and $\tan \beta$ , therefore $\tan \alpha + \tan \beta = 4$ and $\tan \cdot \tan \beta = -9$ .

$\implies \tan(\alpha + \beta)=\frac{4}{10}=\boxed{0.4}$ .