We have,

$5\cos A+3=0\Rightarrow \cos A=-\frac{3}{5}$

Clearly, $A\in \left(90°,180°\right)$

Now, roots of equation

$9x^2+27x+20=0$ 

$\Rightarrow x=\frac{-27\pm \sqrt{729-720}}{18}=\frac{-27\pm 3}{18}$

$\Rightarrow x=-\frac{5}{3}, -\frac{4}{3}$

Then, $-\frac{5}{3}=\sec A$ and $-\frac{4}{3}=\tan A$.

Hence, the roots are $\sec A$ and $\tan A$.