The given quadratic equation is:

\[
5y^2 - 9y - 2 = 0
\]

To solve this, we will use the quadratic formula:

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

Where \(a = 5\), \(b = -9\), and \(c = -2\). Let's plug these values into the formula.

\[
y = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(5)(-2)}}{2(5)}
\]

Now simplify the equation step by step:

\[
y = \frac{9 \pm \sqrt{81 + 40}}{10}
\]
\[
y = \frac{9 \pm \sqrt{121}}{10}
\]
\[
y = \frac{9 \pm 11}{10}
\]

Now, we solve for the two possible values of \(y\):

1. \(y_1 = \frac{9 + 11}{10} = \frac{20}{10} = 2\)
2. \(y_2 = \frac{9 - 11}{10} = \frac{-2}{10} = -0.2\)

Thus, the solutions are:

\[
y = 2 \quad \text{or} \quad y = -0.2
\]

Do you have any questions or would you like more details?

Here are 5 related questions:
1. What is the quadratic formula used for?
2. How do you determine the discriminant in the quadratic formula?
3. How can you check if the roots of a quadratic equation are real or complex?
4. What are other methods to solve quadratic equations besides the quadratic formula?
5. How do quadratic equations apply to real-world problems?

**Tip:** Always double-check the discriminant (\(b^2 - 4ac\)) to ensure that the roots are real, imaginary, or complex.

This page explains how to solve the quadratic equation 5y^2 - 9y - 2 = 0 using the quadratic formula. We provide a detailed, step-by-step breakdown of how to find the roots of the equation, suitable for students in Grades 9-11. Learn how to apply key algebraic concepts such as discriminants and solving quadratic equations.

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