We are given the equation:

\[
11r + 15 = -2r^2
\]

Let's solve this step by step.

### Step 1: Rearrange the equation
Move all terms to one side to set the equation equal to 0:

\[
2r^2 + 11r + 15 = 0
\]

### Step 2: Solve the quadratic equation

Now we have a quadratic equation in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 2 \)
- \( b = 11 \)
- \( c = 15 \)

We will use the quadratic formula:

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

Substitute the values of \(a\), \(b\), and \(c\):

\[
r = \frac{-11 \pm \sqrt{11^2 - 4(2)(15)}}{2(2)}
\]
\[
r = \frac{-11 \pm \sqrt{121 - 120}}{4}
\]
\[
r = \frac{-11 \pm \sqrt{1}}{4}
\]
\[
r = \frac{-11 \pm 1}{4}
\]

### Step 3: Calculate the two possible solutions

1. \( r = \frac{-11 + 1}{4} = \frac{-10}{4} = -2.5 \)
2. \( r = \frac{-11 - 1}{4} = \frac{-12}{4} = -3 \)

### Final answer:
The solutions to the equation are:
\[
r = -2.5 \quad \text{or} \quad r = -3
\]

Would you like more details on any step?

Here are 5 related questions to deepen your understanding:
1. How do you know when to use the quadratic formula?
2. Can all quadratic equations be solved using the quadratic formula?
3. What is the discriminant, and how does it determine the number of solutions in a quadratic equation?
4. Can you explain why quadratic equations have at most two solutions?
5. How would the process change if the quadratic term had a positive coefficient?

**Tip:** Always check the discriminant \( b^2 - 4ac \) before solving, as it helps you determine if real solutions exist!

Learn how to solve the quadratic equation 11r + 15 = -2r² with a step-by-step solution using the quadratic formula. This example covers key algebraic concepts and the discriminant theorem, suitable for students in Grades 8-10.

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