Use the quadratic formula to solve the equation. \[ y^{2}+36=14 y \]

To solve the equation \( y^{2} + 36 = 14y \), first, rearrange it into standard quadratic form: \[ y^{2} - 14y + 36 = 0. \] Now, identify the coefficients: \( a = 1 \), \( b = -14 \), and \( c = 36 \). The quadratic formula is given by \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] Plugging in our coefficients: 1. Calculate \( b^2 - 4ac \): \[ (-14)^2 - 4(1)(36) = 196 - 144 = 52. \] 2. Now, use the quadratic formula: \[ y = \frac{-(-14) \pm \sqrt{52}}{2(1)} = \frac{14 \pm \sqrt{52}}{2}. \] 3. Simplify \( \sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13} \), so: \[ y = \frac{14 \pm 2\sqrt{13}}{2}. \] 4. Finally, divide by 2: \[ y = 7 \pm \sqrt{13}. \] So, the two solutions are: \[ y = 7 + \sqrt{13} \quad \text{and} \quad y = 7 - \sqrt{13}. \]