To find the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this equation, the coefficients are:
- \( a = 1 \)
- \( b = 15 \)
- \( c = 36 \)
First, we calculate the discriminant (\( b^2 - 4ac \)):
\[ b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 \] \[ = 225 - 144 \] \[ = 81 \]
Now that we have the discriminant, we can find the zeros:
\[ x = \frac{-15 \pm \sqrt{81}}{2 \cdot 1} \] \[ = \frac{-15 \pm 9}{2} \]
Now, we will calculate the two possible values for \( x \):
- Using the positive root:
\[ x = \frac{-15 + 9}{2} = \frac{-6}{2} = -3 \]
- Using the negative root:
\[ x = \frac{-15 - 9}{2} = \frac{-24}{2} = -12 \]
Thus, the zeros of the quadratic function are:
\[ x = -3 \quad \text{and} \quad x = -12 \]