We are given the equation:

\[
x - 20 = \frac{260000}{x}
\]

To solve for \( x \), follow these steps:

### Step 1: Eliminate the fraction
Multiply both sides of the equation by \( x \) to eliminate the denominator:

\[
x(x - 20) = 260000
\]

This simplifies to:

\[
x^2 - 20x = 260000
\]

### Step 2: Rearrange the equation
Bring all terms to one side to form a quadratic equation:

\[
x^2 - 20x - 260000 = 0
\]

### Step 3: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:

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

For the equation \(x^2 - 20x - 260000 = 0\), we have:
- \(a = 1\)
- \(b = -20\)
- \(c = -260000\)

Substitute these values into the quadratic formula:

\[
x = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(1)(-260000)}}{2(1)}
\]

Simplify the terms:

\[
x = \frac{20 \pm \sqrt{400 + 1040000}}{2}
\]
\[
x = \frac{20 \pm \sqrt{1040400}}{2}
\]

### Step 4: Calculate the square root and solve for \(x\)

\[
x = \frac{20 \pm 1020}{2}
\]

Now, solve for both possible values of \( x \):

1. \( x = \frac{20 + 1020}{2} = \frac{1040}{2} = 520 \)
2. \( x = \frac{20 - 1020}{2} = \frac{-1000}{2} = -500 \)

### Step 5: Final answer
Thus, the solutions are:

\[
x = 520 \quad \text{or} \quad x = -500
\]

Since \( x \) represents a value that might need to stay positive (depending on context), \( x = 520 \) is often preferred, but both values are valid solutions.

Would you like more details, or do you have any questions?

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Here are 5 related questions:

1. How do we determine when to discard a negative solution in such problems?
2. What are the conditions for using the quadratic formula?
3. Can this quadratic equation be solved by factoring?
4. What is the significance of discriminants in quadratic equations?
5. How can you check your solution to verify its accuracy?

**Tip**: When solving quadratics, always check both solutions by substituting them back into the original equation.

Learn how to solve the quadratic equation x - 20 = 260000 / x using the quadratic formula. This problem involves simplifying a rational equation and applying the quadratic equation formula, suitable for students in Grades 9-12.

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