Let's solve the equation step by step:

We have the equation:

\[
52.3 = 3769.9r^2\left(2 - \frac{1}{r}\right)
\]

### Step 1: Distribute \( r^2 \) on the right-hand side.
First, distribute \( r^2 \) across the terms inside the parentheses:

\[
52.3 = 3769.9r^2 \times 2 - 3769.9r^2 \times \frac{1}{r}
\]

This simplifies to:

\[
52.3 = 7539.8r^2 - 3769.9r
\]

### Step 2: Rearrange the equation.
Bring everything to one side to get a standard quadratic equation form:

\[
7539.8r^2 - 3769.9r - 52.3 = 0
\]

### Step 3: Solve the quadratic equation.
We now have a quadratic equation of the form:

\[
7539.8r^2 - 3769.9r - 52.3 = 0
\]

We can solve this using the quadratic formula:

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

Where:
- \(a = 7539.8\),
- \(b = -3769.9\),
- \(c = -52.3\).

Let's calculate the values for \( r \).The solutions for \( r \) are:

\[
r_1 = 0.5135 \quad \text{and} \quad r_2 = -0.0135
\]

Since \( r \) typically represents a radius or a similar positive quantity in practical applications, we can discard the negative value. Therefore, the solution is:

\[
r = 0.5135
\]

Would you like further details or clarifications?

Here are some related questions you might be interested in:
1. How do we determine when to discard negative solutions in practical problems?
2. What are the steps for solving quadratic equations by factoring?
3. Can this solution be obtained graphically?
4. What are some common applications of quadratic equations in physics?
5. How can we approximate solutions if the quadratic formula is complex?

**Tip:** Always check the discriminant (\(b^2 - 4ac\)) before solving a quadratic equation to ensure real solutions exist.

Learn how to solve the quadratic equation 52.3 = 3769.9r^2(2 - 1/r) with a detailed step-by-step breakdown. This problem explores advanced algebraic concepts, including the quadratic formula. Ideal for students in Grades 10-12.

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