To find the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we can set \( y \) equal to zero and solve for \( x \):
\[ x^2 + 15x + 36 = 0 \]
We can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our quadratic equation, the coefficients are:
- \( a = 1 \)
- \( b = 15 \)
- \( c = 36 \)
Now, we will calculate the discriminant (\( b^2 - 4ac \)):
\[ b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 = 225 - 144 = 81 \]
Since the discriminant is positive, we will have two distinct real zeros.
Now we can substitute the values into the quadratic formula:
\[ x = \frac{-15 \pm \sqrt{81}}{2 \cdot 1} = \frac{-15 \pm 9}{2} \]
This gives us two solutions:
- For the positive square root:
\[ x = \frac{-15 + 9}{2} = \frac{-6}{2} = -3 \]
- For the negative square root:
\[ x = \frac{-15 - 9}{2} = \frac{-24}{2} = -12 \]
Thus, the zeros of the function \( y = x^2 + 15x + 36 \) are:
\[ x = -12 \quad \text{and} \quad x = -3 \]
When listing the smaller zero first, we have:
\[ \boxed{-12} \quad \text{and} \quad \boxed{-3} \]