Answer
There are two possible solutions: $x = 0.7$ in or $x = 9.3$ in
Work Step by Step
According to the description and the figure in the exercise, what they're really asking us is to find $x$ when the area of the rectangle formed ($Area = length \times width)$ mesaures 13 square inches. In other words: $$Area_{gutter} = 13 = (20 - 2x)(x)$$ $$13 = 20x - 2x^2$$ $$0 = -2x^2 + 20x - 13$$ which is a quadratic equation. Here, it would be best to use the Quadratic Formula $x = \frac{-b \frac{+}{} \sqrt{b^2 - 4ac}}{2a}$ to solve for the variable:
$$x = \frac{-(20) \frac{+}{} \sqrt{(20^2 - 4(-2)(-13)}}{2(-2)}$$ $$x = \frac{-20 \frac{+}{} \sqrt{(400 - 104)}}{-4}$$ $$x = \frac{-20 \frac{+}{} \sqrt{296}}{-4}$$ Since the exercise asks us to round to the nearest tenth of an inch, we continue as follows: $$x = \frac{-20 + \sqrt{296}}{-4} \approx 0.7$$ $$OR $$$$x = \frac{-20 - \sqrt{296}}{-4} \approx 9.3$$
