Geometry: Common Core (15th Edition)

$x = 10$
With similar triangles, corresponding sides are proportional. Let's set up a proportion using the two sets of corresponding sides that we are given: $\frac{10}{x + 10} = \frac{x + 2}{x + 14}$ Cross multiply to get rid of the fractions: $(x + 10)(x + 2) = 10(x + 14)$ Multiply the terms out: $x^2 + 10x + 2x + 20 = 10x + 140$ Combine like terms: $x^2 + 12x + 20 = 10x + 140$ Subtract $10x$ from each side of the equation to move variables to the left side of the equation: $x^2 + 2x + 20 = 140$ Subtract $140$ from each side of the equation to move constants to the left side of the equation: $x^2 + 2x - 120 = 0$ Factor the quadratic equation: $(x + 12)(x - 10) = 0$ Set each factor equal to $0$ to solve for $x$: $x + 12 = 0$ Subtract $12$ from each side of the equation to solve: $x = -12$ We discard this solution because $x$ cannot be negative. Let's look at the other factor: $x - 10 = 0$ Add $10$ to each side of the equation: $x = 10$