You have the equation:

\[
y^2 - 20y - 300 = 0
\]

This is a quadratic equation of the form \(ay^2 + by + c = 0\), where \(a = 1\), \(b = -20\), and \(c = -300\).

### To solve this, we can apply the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Let's solve it step-by-step.

1. **Identify coefficients**:  
   \(a = 1\), \(b = -20\), and \(c = -300\)

2. **Calculate the discriminant**:
\[
\Delta = b^2 - 4ac = (-20)^2 - 4(1)(-300) = 400 + 1200 = 1600
\]

3. **Use the quadratic formula**:
\[
y = \frac{-(-20) \pm \sqrt{1600}}{2(1)} = \frac{20 \pm 40}{2}
\]

4. **Solve for \(y\)**:
- \(y_1 = \frac{20 + 40}{2} = \frac{60}{2} = 30\)
- \(y_2 = \frac{20 - 40}{2} = \frac{-20}{2} = -10\)

### Therefore, the solutions are:
\[
y = 30 \quad \text{or} \quad y = -10
\]

Would you like more details or clarifications?

Here are 5 questions for further exploration:
1. How would the quadratic formula change if \(b\) was positive?
2. Can this quadratic equation be solved by factoring?
3. What is the significance of the discriminant in quadratic equations?
4. How would graphing the equation help in visualizing the solutions?
5. What are the applications of quadratic equations in real-world problems?

**Tip**: The discriminant \(b^2 - 4ac\) tells you how many real solutions a quadratic equation has. If it's positive, there are two real solutions!

Learn how to solve the quadratic equation y^2 - 20y - 300 with a step-by-step solution. This algebra problem covers the quadratic formula, discriminants, and solving for real solutions. Ideal for students in Grades 9-10.

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