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To find the value of \( x \) in the given geometric configuration, we can use the properties of right triangles and the Pythagorean theorem.

Given:
- The right triangle with legs \( x \) and \( x + 5 \)
- The hypotenuse of this right triangle is 20

Using the Pythagorean theorem:
\[ a^2 + b^2 = c^2 \]

Here, \( a = x \), \( b = x + 5 \), and \( c = 20 \). Thus, we have:
\[ x^2 + (x + 5)^2 = 20^2 \]

Expanding and simplifying the equation:
\[ x^2 + (x^2 + 10x + 25) = 400 \]
\[ 2x^2 + 10x + 25 = 400 \]
\[ 2x^2 + 10x + 25 - 400 = 0 \]
\[ 2x^2 + 10x - 375 = 0 \]

Dividing the entire equation by 2 to simplify:
\[ x^2 + 5x - 187.5 = 0 \]

This is a quadratic equation of the form \( ax^2 + bx + c = 0 \). We can solve it using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our equation, \( a = 1 \), \( b = 5 \), and \( c = -187.5 \):
\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-187.5)}}{2 \cdot 1} \]
\[ x = \frac{-5 \pm \sqrt{25 + 750}}{2} \]
\[ x = \frac{-5 \pm \sqrt{775}}{2} \]
\[ x = \frac{-5 \pm \sqrt{25 \cdot 31}}{2} \]
\[ x = \frac{-5 \pm 5\sqrt{31}}{2} \]
\[ x = \frac{-5 + 5\sqrt{31}}{2} \quad \text{or} \quad x = \frac{-5 - 5\sqrt{31}}{2} \]

Since \( x \) represents a length, we take the positive value:
\[ x = \frac{-5 + 5\sqrt{31}}{2} \]

Simplifying further:
\[ x = \frac{5(\sqrt{31} - 1)}{2} \]

Thus, the value of \( x \) is:
\[ x \approx 6.1 \]

Would you like further details or have any questions?

Here are five more questions you can ask:
1. How do I solve another quadratic equation?
2. Can you explain the steps of the quadratic formula?
3. What are the properties of right triangles?
4. How is the Pythagorean theorem used in different types of problems?
5. How can I verify the solution to a geometric problem?

**Tip:** Always double-check your algebraic manipulations when solving equations to avoid small errors that can lead to incorrect solutions.

Learn how to find the value of x in a right triangle where one leg is x, the other leg is x + 5, and the hypotenuse is 20. This problem involves using the Pythagorean theorem and solving a quadratic equation. Suitable for Grades 9-12 students.

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