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I see the assignment involves solving quadratic equations using three different methods: factoring, completing the square, and the quadratic formula. Let's go over each problem and how to solve them according to the specified methods.

### A. **Factoring**
1. **\( 3x - x^2 = 0 \)**
   - Rearrange the equation: \( -x^2 + 3x = 0 \), or better yet, \( x(3 - x) = 0 \).
   - Set each factor equal to zero:  
     - \( x = 0 \)
     - \( 3 - x = 0 \) → \( x = 3 \)
   - Solutions: \( x = 0 \) or \( x = 3 \).

2. **\( x^2 + 2x - 3 = 0 \)**
   - This is a standard quadratic equation that can be factored:  
     \( x^2 + 2x - 3 = (x + 3)(x - 1) = 0 \).
   - Set each factor equal to zero:  
     - \( x + 3 = 0 \) → \( x = -3 \)
     - \( x - 1 = 0 \) → \( x = 1 \)
   - Solutions: \( x = -3 \) or \( x = 1 \).

---

### B. **Completing the Square**
1. **\( x^2 - 6x = 8 \)**
   - First, move the constant to the other side:  
     \( x^2 - 6x - 8 = 0 \) → \( x^2 - 6x = 8 \).
   - To complete the square, take half of the coefficient of \( x \), square it, and add it to both sides.  
     Half of \(-6\) is \(-3\), and \((-3)^2 = 9\):  
     \( x^2 - 6x + 9 = 8 + 9 \), which gives:  
     \( (x - 3)^2 = 17 \).
   - Take the square root of both sides:  
     \( x - 3 = \pm \sqrt{17} \).
   - Solutions: \( x = 3 \pm \sqrt{17} \).

2. **\( 2x^2 + 8x = 42 \)**
   - Divide the entire equation by 2:  
     \( x^2 + 4x = 21 \).
   - Take half of the coefficient of \( x \), square it, and add to both sides. Half of \( 4 \) is \( 2 \), and \( 2^2 = 4 \):  
     \( x^2 + 4x + 4 = 21 + 4 \), which gives:  
     \( (x + 2)^2 = 25 \).
   - Take the square root of both sides:  
     \( x + 2 = \pm 5 \).
   - Solutions: \( x = -2 \pm 5 \), which gives \( x = 3 \) or \( x = -7 \).

---

### C. **Quadratic Formula**
The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( ax^2 + bx + c = 0 \).

1. **\( x^2 - 121 = 0 \)**
   - Rearrange: \( x^2 = 121 \).
   - Taking the square root of both sides:  
     \( x = \pm \sqrt{121} \).
   - Solution: \( x = 11 \) or \( x = -11 \).

2. **\( 12x^2 - 73x + 6 = 0 \)**
   - Here, \( a = 12 \), \( b = -73 \), and \( c = 6 \).
   - Apply the quadratic formula:  
     \[
     x = \frac{-(-73) \pm \sqrt{(-73)^2 - 4(12)(6)}}{2(12)}
     \]
     Simplifying:
     \[
     x = \frac{73 \pm \sqrt{5329 - 288}}{24} = \frac{73 \pm \sqrt{5041}}{24} = \frac{73 \pm 71}{24}
     \]
   - This gives two solutions:  
     \[
     x = \frac{73 + 71}{24} = \frac{144}{24} = 6
     \]
     and  
     \[
     x = \frac{73 - 71}{24} = \frac{2}{24} = \frac{1}{12}.
     \]
   - Solutions: \( x = 6 \) or \( x = \frac{1}{12} \).

---

Would you like further explanation on any of these methods or problems? Here are some related questions you might explore:

1. How can factoring be used in more complex quadratic equations?
2. What are the benefits of completing the square over other methods?
3. When is it ideal to use the quadratic formula versus other methods?
4. What happens when the discriminant in the quadratic formula is negative?
5. Can completing the square be used for solving cubic or higher-degree polynomials?

**Tip**: Always check your solutions by plugging them back into the original equation!

Solve the quadratic equations using the specified methods: factoring, completing the square, and the quadratic formula.

Learn how to solve quadratic equations using three key methods: factoring, completing the square, and the quadratic formula. This guide provides step-by-step solutions for each method, helping students in Grades 8-10 understand the algebraic concepts and formulas involved.

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