Determine the zeros of the quadratic function y=x2+15x+36. (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point) smaller zero: x=; greater zero: x=

To find the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we can use the quadratic formula, which is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In this case, 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, there are two distinct real zeros.

Next, we will apply the quadratic formula:

\[
x = \frac{-15 \pm \sqrt{81}}{2 \cdot 1}
\]
\[
= \frac{-15 \pm 9}{2}
\]

Now we will calculate the two possible values for \( x \):

1. For the plus sign:
\[
x = \frac{-15 + 9}{2} = \frac{-6}{2} = -3
\]

2. For the minus sign:
\[
x = \frac{-15 - 9}{2} = \frac{-24}{2} = -12
\]

Now we can list the zeros. The smaller zero is \( -12 \) and the greater zero is \( -3 \).

So, we have:
- smaller zero: \( x = -12 \)
- greater zero: \( x = -3 \)