To solve the equation \((6.2 - x)(3.9 - x) = 8.38\), we need to expand and simplify the expression, then solve for \(x\).

### Step 1: Expand the equation
\[
(6.2 - x)(3.9 - x) = 8.38
\]
Apply the distributive property (FOIL method):

\[
6.2 \times 3.9 - 6.2 \times x - 3.9 \times x + x^2 = 8.38
\]
Simplify the multiplication:
\[
24.18 - 6.2x - 3.9x + x^2 = 8.38
\]

### Step 2: Combine like terms
\[
x^2 - 10.1x + 24.18 = 8.38
\]

### Step 3: Move all terms to one side
\[
x^2 - 10.1x + 24.18 - 8.38 = 0
\]
Simplify:
\[
x^2 - 10.1x + 15.8 = 0
\]

### Step 4: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For \(x^2 - 10.1x + 15.8 = 0\), the coefficients are:
- \(a = 1\)
- \(b = -10.1\)
- \(c = 15.8\)

Substitute these values into the quadratic formula:

\[
x = \frac{-(-10.1) \pm \sqrt{(-10.1)^2 - 4(1)(15.8)}}{2(1)}
\]

Simplify:
\[
x = \frac{10.1 \pm \sqrt{102.01 - 63.2}}{2}
\]
\[
x = \frac{10.1 \pm \sqrt{38.81}}{2}
\]
\[
x = \frac{10.1 \pm 6.23}{2}
\]

### Step 5: Find the two possible solutions
1. \(x = \frac{10.1 + 6.23}{2} = \frac{16.33}{2} = 8.165\)
2. \(x = \frac{10.1 - 6.23}{2} = \frac{3.87}{2} = 1.935\)

### Final Solution:
The two possible values for \(x\) are:
\[
x = 8.165 \quad \text{or} \quad x = 1.935
\]

Would you like more details on any step?

Here are five related questions:
1. How can we graph the equation to verify the solution?
2. What other methods can be used to solve quadratic equations?
3. How does the discriminant affect the number of solutions in a quadratic equation?
4. What real-world scenarios involve solving quadratic equations like this one?
5. How do you factor a quadratic equation?

**Tip:** When using the quadratic formula, always double-check the discriminant (\(b^2 - 4ac\)) to determine if real solutions exist!

Learn how to solve the quadratic equation (6.2 - x)(3.9 - x) = 8.38 through step-by-step expansion, combining like terms, and applying the quadratic formula. This problem introduces key concepts in algebra, suitable for students in Grades 9-11.

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