Continued Fraction Evaluation

We can express $y$ recursively as: $y = \frac{1}{2 + \frac{1}{3 + y}} = \frac{3 + y}{2(3 + y) + 1} = \frac{3 + y}{7 + 2y}$ $y(7 + 2y) = 3 + y \implies 2y^2 + 6y - 3 = 0$ Using the quadratic formula: $y = \frac{-6 \pm \sqrt{36 - 4(2)(-3)}}{4} = \frac{-6 \pm \sqrt{60}}{4} = \frac{-3 \pm \sqrt{15}}{2}$ Since $y > 0$, $y = \frac{\sqrt{15} - 3}{2}$.