If $$\sin x = -\frac{3}{5}$$, where $$\pi < x < \frac{3\pi}{2}$$, then $$80(\tan^2 x - \cos x)$$ is equal to

Given that $$\sin x = -\frac{3}{5}$$ and $$\pi < x < \frac{3\pi}{2}$$, which is the third quadrant. In this quadrant, sine and cosine are negative, and tangent is positive.

Using the Pythagorean identity, $$\sin^2 x + \cos^2 x = 1$$:

Substitute $$\sin x = -\frac{3}{5}$$:

$$\left(-\frac{3}{5}\right)^2 + \cos^2 x = 1$$

$$\frac{9}{25} + \cos^2 x = 1$$

$$\cos^2 x = 1 - \frac{9}{25} = \frac{16}{25}$$

Thus, $$\cos x = \pm \frac{4}{5}$$.

Since $$x$$ is in the third quadrant, cosine is negative, so $$\cos x = -\frac{4}{5}$$.

Now, find $$\tan x$$:

$$\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{3}{5}}{-\frac{4}{5}} = \frac{3}{4}$$.

Compute $$\tan^2 x - \cos x$$:

$$\tan^2 x = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$$

$$\cos x = -\frac{4}{5}$$

So, $$\tan^2 x - \cos x = \frac{9}{16} - \left(-\frac{4}{5}\right) = \frac{9}{16} + \frac{4}{5}$$

Add the fractions. The least common denominator of 16 and 5 is 80:

$$\frac{9}{16} = \frac{9 \times 5}{16 \times 5} = \frac{45}{80}$$

$$\frac{4}{5} = \frac{4 \times 16}{5 \times 16} = \frac{64}{80}$$

Thus, $$\frac{45}{80} + \frac{64}{80} = \frac{109}{80}$$

Now, compute $$80 \times (\tan^2 x - \cos x) = 80 \times \frac{109}{80} = 109$$.

The expression $$80(\tan^2 x - \cos x)$$ equals 109, which corresponds to option B.