The problem asks us to simplify the equation $\frac{1}{18} + \frac{1}{y} = \frac{2}{y+3}$ and then solve it using the quadratic formula.

1. Problem Description

The problem asks us to simplify the equation $\frac{1}{18} + \frac{1}{y} = \frac{2}{y+3}$ and then solve it using the quadratic formula.

2. Solution Steps

First, we clear the denominators by multiplying both sides of the equation by $18y(y+3)$:

$18y(y+3) \cdot (\frac{1}{18} + \frac{1}{y}) = 18y(y+3) \cdot \frac{2}{y+3}$

$y(y+3) + 18(y+3) = 36y$

$y^2 + 3y + 18y + 54 = 36y$

$y^2 + 21y + 54 = 36y$

$y^2 + 21y - 36y + 54 = 0$

$y^2 - 15y + 54 = 0$

Now, we use the quadratic formula to solve for $y$. The quadratic formula is given by

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a=1$, $b=-15$, and $c=54$.

$y = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(1)(54)}}{2(1)}$

$y = \frac{15 \pm \sqrt{225 - 216}}{2}$

$y = \frac{15 \pm \sqrt{9}}{2}$

$y = \frac{15 \pm 3}{2}$

So we have two possible solutions:

$y = \frac{15 + 3}{2} = \frac{18}{2} = 9$

$y = \frac{15 - 3}{2} = \frac{12}{2} = 6$

3. Final Answer

The solutions are $y = 6, 9$.