If an angle A of a $$\triangle ABC$$ satisfies $$5\cos A + 3 = 0$$, then the roots of the quadratic equation $$9x^2 + 27x + 20 = 0$$ are:

We are told that in $$\triangle ABC$$ the angle $$A$$ satisfies the relation $$5\cos A + 3 = 0$$.

First, we isolate $$\cos A$$:

$$5\cos A + 3 = 0 \;\Longrightarrow\; 5\cos A = -3 \;\Longrightarrow\; \cos A = -\dfrac{3}{5}.$$

Since $$A$$ is an angle of a triangle, we know $$0 < A < \pi$$. The value $$\cos A = -\dfrac{3}{5}$$ is negative, so $$A$$ lies in the second quadrant, where sine is positive and cosine is negative.

Now we compute $$\sin A$$ using the fundamental Pythagorean identity $$\sin^2\!A + \cos^2\!A = 1$$.

$$\sin^2\!A = 1 - \cos^2\!A = 1 - \left(-\dfrac{3}{5}\right)^2 = 1 - \dfrac{9}{25} = \dfrac{25}{25} - \dfrac{9}{25} = \dfrac{16}{25}.$$

Because $$A$$ is in the second quadrant, $$\sin A$$ is positive, hence

$$\sin A = \dfrac{4}{5}.$$

Next we find the other required trigonometric ratios.

The secant is the reciprocal of cosine, so

$$\sec A = \dfrac{1}{\cos A} = \dfrac{1}{-\dfrac{3}{5}} = -\dfrac{5}{3}.$$

The tangent is the quotient of sine by cosine, so

$$\tan A = \dfrac{\sin A}{\cos A} = \dfrac{\dfrac{4}{5}}{-\dfrac{3}{5}} = -\dfrac{4}{3}.$$

Now we examine the quadratic equation

$$9x^2 + 27x + 20 = 0$$

and test whether $$x = \sec A$$ or $$x = \tan A$$ satisfy it.

Substituting $$x = \sec A = -\dfrac{5}{3}$$:

$$9\left(-\dfrac{5}{3}\right)^2 + 27\left(-\dfrac{5}{3}\right) + 20 = 9\left(\dfrac{25}{9}\right) + 27\left(-\dfrac{5}{3}\right) + 20 = 25 - 45 + 20 = 0.$$

So $$x = \sec A$$ is indeed a root.

Substituting $$x = \tan A = -\dfrac{4}{3}$$:

$$9\left(-\dfrac{4}{3}\right)^2 + 27\left(-\dfrac{4}{3}\right) + 20 = 9\left(\dfrac{16}{9}\right) + 27\left(-\dfrac{4}{3}\right) + 20 = 16 - 36 + 20 = 0.$$

Thus $$x = \tan A$$ is also a root.

Since the two roots of the quadratic are $$\sec A$$ and $$\tan A$$, they correspond exactly to Option B.

Hence, the correct answer is Option B.