\(\dfrac{1}{2x} = \dfrac{r-x}{7}\\ 7 = 2rx - 2x^2\\ 2x^2 - 2rx + 7 = 0\\ \text{There will be exactly one real solution if the discriminant is equal to 0}\\ D = (-2r)^2 - 4(2)(7) = 0\\ 4r^2 = 56\\ r^2 = 14\\ r = \pm \sqrt{14}\\ \sqrt{14}(-\sqrt{14}) = -14\)