The problem is to solve the equation $\frac{1}{12} + \frac{1}{y} = \frac{3}{y+6}$ for $y$ using the quadratic formula.

1. Problem Description

The problem is to solve the equation $\frac{1}{12} + \frac{1}{y} = \frac{3}{y+6}$ for $y$ using the quadratic formula.

2. Solution Steps

First, we clear the fractions by multiplying both sides of the equation by $12y(y+6)$:

$12y(y+6) \left(\frac{1}{12} + \frac{1}{y}\right) = 12y(y+6) \left(\frac{3}{y+6}\right)$

$y(y+6) + 12(y+6) = 36y$

$y^2 + 6y + 12y + 72 = 36y$

$y^2 + 18y + 72 = 36y$

$y^2 - 18y + 72 = 0$

Now we use the quadratic formula to solve for $y$:

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

where $a=1$, $b=-18$, and $c=72$.

$y = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(1)(72)}}{2(1)}$

$y = \frac{18 \pm \sqrt{324 - 288}}{2}$

$y = \frac{18 \pm \sqrt{36}}{2}$

$y = \frac{18 \pm 6}{2}$

$y_1 = \frac{18 + 6}{2} = \frac{24}{2} = 12$

$y_2 = \frac{18 - 6}{2} = \frac{12}{2} = 6$

Therefore, the solutions are $y = 6$ and $y = 12$.

3. Final Answer

$6, 12$