Let's solve the given quadratic equations based on the examples from the green board (Factoring and Completing the Square methods). We will use the appropriate method for each problem.

1. $x^2 + 10x + 9 = 0$

   We can solve this by factoring. We need to find two numbers that multiply to 9 and add to 10. The numbers are 1 and 9.

   $(x + 1)(x + 9) = 0$

   Therefore, the solutions are: $x = -1 \quad \text{or} \quad x = -9$

3. $x^2 + 5x - 14 = 0$

   To solve this, we apply factoring. We need two numbers that multiply to $-14$ and add to 5. Unfortunately, this does not factor easily, so we will use the quadratic formula:

   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 5$, and $c = -14$.

   $x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-14)}}{2(1)}$ $x = \frac{-5 \pm \sqrt{25 + 56}}{2}$ $x = \frac{-5 \pm \sqrt{81}}{2}$ $x = \frac{-5 \pm 9}{2}$

   So, the solutions are: $x = \frac{-5 + 9}{2} = 2 \quad \text{or} \quad x = \frac{-5 - 9}{2} = -7$

5. $x^2 + 7x = 4$

   First, rearrange the equation: $x^2 + 7x - 4 = 0$

   We will apply the quadratic formula since factoring isn't straightforward.

   $x = \frac{-7 \pm \sqrt{7^2 - 4(1)(-4)}}{2(1)}$ $x = \frac{-7 \pm \sqrt{49 + 16}}{2}$ $x = \frac{-7 \pm \sqrt{65}}{2}$

   Therefore, the solutions are: $x = \frac{-7 \pm \sqrt{65}}{2}$ which gives two solutions in exact form.

7. $4x^2 - 4x + 1 = 0$

   This is a perfect square trinomial. We can write it as: $(2x - 1)^2 = 0$

   So, the solution is: $x = \frac{1}{2}$

9. $9x^2 - 72 = 0$

   Start by factoring: $9(x^2 - 8) = 0$

   Then: $x^2 - 8 = 0$ $x^2 = 8$

   Taking the square root of both sides: $x = \pm \sqrt{8} = \pm 2\sqrt{2}$

Do you have any questions or want more details on any step? Here are 5 related problems for practice:

1. Solve $x^2 - 6x + 9 = 0$ using factoring.
2. Solve $x^2 - 7x + 12 = 0$ by completing the square.
3. Solve $3x^2 - 4x + 2 = 0$ using the quadratic formula.
4. Factor and solve $x^2 - 9x + 20 = 0$.
5. Use the quadratic formula to solve $x^2 + 3x - 10 = 0$.

Tip: Always check if a quadratic can be factored before using the quadratic formula—it can save time!