To solve the quadratic equation \( y^2 + 12y + 27 = 0 \), we can use the quadratic formula:

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

For the given equation:

- \( a = 1 \)
- \( b = 12 \)
- \( c = 27 \)

Now, let's plug in the values into the formula:

\[
y = \frac{-12 \pm \sqrt{12^2 - 4(1)(27)}}{2(1)}
\]
\[
y = \frac{-12 \pm \sqrt{144 - 108}}{2}
\]
\[
y = \frac{-12 \pm \sqrt{36}}{2}
\]
\[
y = \frac{-12 \pm 6}{2}
\]

This gives us two possible solutions:

1. \( y = \frac{-12 + 6}{2} = \frac{-6}{2} = -3 \)
2. \( y = \frac{-12 - 6}{2} = \frac{-18}{2} = -9 \)

So, the solutions are \( y = -3 \) and \( y = -9 \).

Would you like further details on this solution, or do you have any questions?

Here are five related questions to consider:

1. How would you factorize the quadratic equation instead of using the quadratic formula?
2. Can the discriminant be used to determine the nature of the roots for this quadratic equation?
3. What happens if we change the constant term in the equation to a different number?
4. How can you graph the quadratic equation \( y^2 + 12y + 27 = 0 \)?
5. What are the applications of solving quadratic equations in real-life situations?

**Tip:** Always check the discriminant \( b^2 - 4ac \) before solving a quadratic equation to understand the nature of the roots (real and distinct, real and equal, or complex).

Discover how to solve the quadratic equation y² + 12y + 27 = 0 using the quadratic formula. This problem is ideal for students in Grades 9-10, covering essential algebraic concepts and demonstrating the use of the discriminant to determine the nature of the roots.

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