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### Problem 1:
Solve \( x^2 + (4 - k)x + (k - 6) = 0 \) for \( k \).

**Solution:**

1. To solve for \( x \) in terms of \( k \), we use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
2. For the given equation, \( a = 1 \), \( b = 4 - k \), and \( c = k - 6 \).
3. Substitute these values into the quadratic formula:
   \[
   x = \frac{-(4 - k) \pm \sqrt{(4 - k)^2 - 4 \cdot 1 \cdot (k - 6)}}{2 \cdot 1}
   \]
4. Simplify the expression inside the square root:
   \[
   x = \frac{-(4 - k) \pm \sqrt{(4 - k)^2 - 4(k - 6)}}{2}
   \]
   \[
   x = \frac{-(4 - k) \pm \sqrt{16 - 8k + k^2 - 4k + 24}}{2}
   \]
   \[
   x = \frac{-(4 - k) \pm \sqrt{k^2 - 12k + 40}}{2}
   \]

### Problem 2:
Find the range of values of \( k \) for which both roots of the equation \( x^2 + (4 - k)x + (k - 6) = 0 \) are real.

**Solution:**

1. The roots of the quadratic equation are real if the discriminant is non-negative.
2. The discriminant \( \Delta \) is given by:
   \[
   \Delta = b^2 - 4ac
   \]
3. For our equation \( x^2 + (4 - k)x + (k - 6) = 0 \), \( a = 1 \), \( b = 4 - k \), and \( c = k - 6 \).
4. Therefore, the discriminant is:
   \[
   \Delta = (4 - k)^2 - 4 \cdot 1 \cdot (k - 6)
   \]
   \[
   \Delta = 16 - 8k + k^2 - 4k + 24
   \]
   \[
   \Delta = k^2 - 12k + 40
   \]
5. For the roots to be real, \( \Delta \geq 0 \):
   \[
   k^2 - 12k + 40 \geq 0
   \]
6. We solve the inequality \( k^2 - 12k + 40 \geq 0 \) by finding the roots of the equation:
   \[
   k^2 - 12k + 40 = 0
   \]
   Using the quadratic formula:
   \[
   k = \frac{12 \pm \sqrt{144 - 160}}{2}
   \]
   \[
   k = \frac{12 \pm \sqrt{-16}}{2}
   \]
   Since the discriminant is negative, \( k^2 - 12k + 40 \) does not factor into real numbers and is always positive for real \( k \). Therefore, there are no restrictions on \( k \) for the roots to be real.

### Problem 3:
Determine the conditions under which the quadratic equation has equal roots.

**Solution:**

1. The quadratic equation has equal roots when the discriminant is zero.
2. Set the discriminant equal to zero:
   \[
   (4 - k)^2 - 4(k - 6) = 0
   \]
   Simplify:
   \[
   16 - 8k + k^2 - 4k + 24 = 0
   \]
   \[
   k^2 - 12k + 40 = 0
   \]
3. Solve for \( k \):
   \[
   k = \frac{12 \pm \sqrt{144 - 160}}{2}
   \]
   \[
   k = \frac{12 \pm \sqrt{-16}}{2}
   \]
   The discriminant is negative, indicating no real roots. Therefore, there are no real values of \( k \) that make the roots equal.

Would you like any additional details or have any questions about these solutions?

### Follow-up Questions:

1. What are the steps to derive the quadratic formula?
2. How do you determine the nature of the roots using the discriminant?
3. What is the significance of the coefficients in a quadratic equation?
4. How do you factorize a quadratic equation?
5. What is the role of the vertex in the graph of a quadratic function?
6. How do you complete the square for a quadratic equation?
7. What are the applications of quadratic equations in real life?
8. How does changing the value of \( k \) affect the graph of the quadratic function?

### Tip:
When solving quadratic equations, always check the discriminant first to understand the nature of the roots.

Learn how to solve the quadratic equation x^2 + (4 - k)x + (k - 6) = 0 for k with detailed step-by-step solutions. This problem involves quadratic equations, discriminants, and the quadratic formula theorem. Suitable for advanced high school students in Grades 11-12.

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