To solve the quadratic equation \( 6y^2 + 6y + 1 = 0 \), we can use the quadratic formula:

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

For the equation \( 6y^2 + 6y + 1 = 0 \), we have:
- \( a = 6 \)
- \( b = 6 \)
- \( c = 1 \)

Now, let's plug these values into the quadratic formula:

\[
y = \frac{-6 \pm \sqrt{6^2 - 4(6)(1)}}{2(6)}
\]

This simplifies to:

\[
y = \frac{-6 \pm \sqrt{36 - 24}}{12}
\]

\[
y = \frac{-6 \pm \sqrt{12}}{12}
\]

\[
y = \frac{-6 \pm 2\sqrt{3}}{12}
\]

Now simplify the fraction:

\[
y = \frac{-6}{12} \pm \frac{2\sqrt{3}}{12}
\]

\[
y = -\frac{1}{2} \pm \frac{\sqrt{3}}{6}
\]

Thus, the two solutions are:

\[
y_1 = -\frac{1}{2} + \frac{\sqrt{3}}{6}, \quad y_2 = -\frac{1}{2} - \frac{\sqrt{3}}{6}
\]

Would you like any further clarification or details?

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Here are 5 related questions to expand on this:

1. How would you factor a quadratic equation like this one instead of using the quadratic formula?
2. Can you graph the solutions of this quadratic equation?
3. What are the steps to derive the quadratic formula from a general quadratic equation?
4. How do complex solutions differ from real solutions in quadratic equations?
5. How can the discriminant \( b^2 - 4ac \) tell you about the number and type of solutions?

**Tip:** Always check the discriminant (\( b^2 - 4ac \)) before solving—it helps you know how many and what kind of solutions to expect.

This detailed solution demonstrates how to solve the quadratic equation 6y² + 6y + 1 using the quadratic formula. Learn about key concepts such as the discriminant and quadratic solutions, suitable for students in Grades 9-10.

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