You have uploaded two images containing several polynomial equations. Let’s solve the equations from both images.

First image:

1. $x + 1 = 0$
   Solution: $x = -1$

2. $x - 5 = 0$
   Solution: $x = 5$

3. $2x - 4 = 0$
   Solution: $x = \frac{4}{2} = 2$

4. $10x + 5 = 0$
   Solution: $10x = -5$ → $x = -\frac{1}{2}$

5. $3x - 7 = 0$
   Solution: $x = \frac{7}{3}$

Second image:

6. $x^2 - 9 = 0$
   This is a difference of squares:
   $(x - 3)(x + 3) = 0$
   Solutions: $x = 3$ or $x = -3$

7. $x^2 - 4 = 0$
   Another difference of squares:
   $(x - 2)(x + 2) = 0$
   Solutions: $x = 2$ or $x = -2$

8. $x^2 + 9 = 0$
   There are no real solutions since $x^2 + 9$ is always positive.
   Solution: No real solution.

9. $x(x - 1) = 0$
   Solutions: $x = 0$ or $x = 1$

10. $(x - 2)(x + 3) = 0$
    Solutions: $x = 2$ or $x = -3$

11. $x(x^2 + 1) = 0$
    Solutions: $x = 0$ (since $x^2 + 1$ has no real roots).

12. $2x^2 + 2x = 0$
    Factor out $2x$:
    $2x(x + 1) = 0$
    Solutions: $x = 0$ or $x = -1$

13. $x^2 - x = 0$
    Factor:
    $x(x - 1) = 0$
    Solutions: $x = 0$ or $x = 1$

14. $2x - x^2 + 3x - 3 = 0$
    Combine like terms:
    $-x^2 + 5x - 3 = 0$
    This is a quadratic equation. Use the quadratic formula:
    $x = \frac{-5 \pm \sqrt{5^2 - 4(-1)(-3)}}{2(-1)}$
    Simplifying:
    $x = \frac{-5 \pm \sqrt{25 - 12}}{-2} = \frac{-5 \pm \sqrt{13}}{-2}$
    Solutions: $x = \frac{5 \pm \sqrt{13}}{2}$

Would you like more details or explanations for any specific equation?

Here are 5 related questions to explore further:

1. What is the quadratic formula, and how is it derived?
2. How do you factor polynomials using the difference of squares method?
3. What are imaginary numbers, and why do we use them for solutions like $x^2 + 9 = 0$?
4. How do we solve quadratic equations when the discriminant is negative?
5. What are the common techniques for factoring polynomials?

Tip: When solving quadratic equations, always check if they can be factored easily before using the quadratic formula.