Solve (16)/(1-5x) + (6)/(x) = -1

The common denominator for $\frac{16}{1-5x}$ and $\frac{6}{x}$ is $x(1-5x)$.
Multiply each fraction by the appropriate term to get the common denominator:

$$\frac{16}{1-5x} \cdot \frac{x}{x} + \frac{6}{x} \cdot \frac{1-5x}{1-5x} = -1$$ $$\frac{16x}{x(1-5x)} + \frac{6(1-5x)}{x(1-5x)} = -1$$

$$\frac{16x + 6(1-5x)}{x(1-5x)} = -1$$ $$\frac{16x + 6 - 30x}{x - 5x^2} = -1$$ $$\frac{6 - 14x}{x - 5x^2} = -1$$

$$6 - 14x = -1(x - 5x^2)$$ $$6 - 14x = -x + 5x^2$$

Move all terms to one side of the equation:

$$0 = 5x^2 - x + 14x - 6$$ $$0 = 5x^2 + 13x - 6$$

For $5x^2 + 13x - 6 = 0$, we have $a=5$, $b=13$, $c=-6$.
First, calculate the discriminant $\Delta = b^2 - 4ac$:

$$\Delta = (13)^2 - 4(5)(-6)$$ $$\Delta = 169 - (-120)$$ $$\Delta = 169 + 120$$ $$\Delta = 289$$

Now, substitute the values into the quadratic formula:

$$x = \frac{-13 \pm \sqrt{289}}{2(5)}$$ $$x = \frac{-13 \pm 17}{10}$$

$$x_1 = \frac{-13 + 17}{10} = \frac{4}{10} = \frac{2}{5}$$ $$x_2 = \frac{-13 - 17}{10} = \frac{-30}{10} = -3$$

Both solutions are valid as they do not make the original denominators zero ($x \neq 0$ and $1-5x \neq 0 \implies x \neq \frac{1}{5}$).

The solutions are $x = \frac{2}{5}, -3$.