https://file.aiquickdraw.com/mathbot/file/1729256036167i9043w9e.jpg

The image contains a set of mathematical equations, most of which are quadratic equations. I'll go through them step by step:

1. **\(x^2 + 7x + 12 = 0\)**
   - This is a standard quadratic equation. We can factor it as:
     \[
     (x + 3)(x + 4) = 0
     \]
     So, the solutions are \(x = -3\) and \(x = -4\).

2. **\(x^2 + 11x + 28 = 0\)**
   - Factor the quadratic:
     \[
     (x + 4)(x + 7) = 0
     \]
     Solutions: \(x = -4\) and \(x = -7\).

3. **\(x^2 - 14x + 49 = 0\)**
   - This is a perfect square:
     \[
     (x - 7)^2 = 0
     \]
     Solution: \(x = 7\).

4. **\(x^2 - 12x + 21 = 0\)**
   - Use the quadratic formula here:
     \[
     x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(21)}}{2(1)}
     \]
     Simplifying:
     \[
     x = \frac{12 \pm \sqrt{144 - 84}}{2} = \frac{12 \pm \sqrt{60}}{2} = \frac{12 \pm 2\sqrt{15}}{2} = 6 \pm \sqrt{15}
     \]
     Solutions: \(x = 6 + \sqrt{15}\) and \(x = 6 - \sqrt{15}\).

5. **\(x^2 + 8x - 20 = 0\)**
   - Use the quadratic formula:
     \[
     x = \frac{-8 \pm \sqrt{8^2 - 4(1)(-20)}}{2(1)} = \frac{-8 \pm \sqrt{64 + 80}}{2} = \frac{-8 \pm \sqrt{144}}{2} = \frac{-8 \pm 12}{2}
     \]
     Solutions: \(x = 2\) and \(x = -10\).

6. **\(y^2 + 9y - 36 = 0\)**
   - Factor:
     \[
     (y + 12)(y - 3) = 0
     \]
     Solutions: \(y = -12\) and \(y = 3\).

7. **\(18 + 6y - y^2 = 0\)**
   - Rearrange it as:
     \[
     y^2 - 6y - 18 = 0
     \]
     Use the quadratic formula:
     \[
     y = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-18)}}{2(1)} = \frac{6 \pm \sqrt{36 + 72}}{2} = \frac{6 \pm \sqrt{108}}{2} = \frac{6 \pm 6\sqrt{3}}{2}
     \]
     Simplifying:
     \[
     y = 3 \pm 3\sqrt{3}
     \]
     Solutions: \(y = 3 + 3\sqrt{3}\) and \(y = 3 - 3\sqrt{3}\).

8. **\(z^2 + 9y - y^2 = 0\)**
   - This is not factorable easily, so you might need to apply the quadratic formula again, but let me know if you need this in more detail.

9. **\(x(y - 11) - 4z = 0\)**
   - Solve for \(x\) in terms of \(y\) and \(z\):
     \[
     x(y - 11) = 4z \quad \Rightarrow \quad x = \frac{4z}{y - 11}
     \]

10. **\(y^2 - 4(2y - 3) = 0\)**
    - Expand and simplify:
      \[
      y^2 - 8y + 12 = 0
      \]
      Solve using the quadratic formula:
      \[
      y = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(12)}}{2(1)} = \frac{8 \pm \sqrt{64 - 48}}{2} = \frac{8 \pm \sqrt{16}}{2} = \frac{8 \pm 4}{2}
      \]
      Solutions: \(y = 6\) and \(y = 2\).

Would you like more details on any of these solutions?

Here are 5 questions for further exploration:

1. How do you apply the quadratic formula in different types of problems?
2. How can we verify whether a quadratic equation is factorable?
3. What is the significance of the discriminant in determining the nature of solutions for quadratic equations?
4. How does completing the square compare to using the quadratic formula for solving quadratics?
5. How can factoring techniques help simplify more complex algebraic expressions?

**Tip**: When solving quadratics, always check if the equation can be factored easily before using the quadratic formula; this can save time!

Solve the following quadratic and algebraic equations: x^2 + 7x + 12 = 0, x^2 + 11x + 28 = 0, x^2 - 14x + 49 = 0, x^2 - 12x + 21 = 0, x^2 + 8x - 20 = 0, y^2 + 9y - 36 = 0, 18 + 6y - y^2 = 0, z^2 + 9y - y^2 = 0, x(y - 11) - 4z = 0, y^2 - 4(2y - 3) = 0.

This page provides detailed solutions to quadratic and algebraic equations, including factoring techniques and using the quadratic formula. Equations include x^2 + 7x + 12 = 0, x^2 + 11x + 28 = 0, and more. Suitable for students in Grades 9-11.

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