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We can solve by factoring: x^2 - 2x + 1 = 0. Let's find two numbers which product is 1 and algebraic addition is -2, so the numbers would be: -1 and -1 since (-1)(-1)= 1 and -1-1 = -2. Then it would factor like this:

x^2 - 2x + 1 = 0 -------> (x - 1)(x - 1) = 0  -------> (x - 1)^2 = 0 ---------> x - 1 = 0  ------> x - 1 + 1 = 0 + 1  ----> x = 1

We do something similar with the other exercises.

x^2 + 9x + 20 = 0  -------->  (x + 4)(x + 5) = 0  --------> x + 4 = 0  and   x + 5 = 0 --------> x = - 4 and x = - 5

3x^2 - 5x - 12 = 0 . Find two numbers that their product give - 36 (it comes from 3 by -12) and their addition give     - 5. So the numbers would be:  -9 and 4 since (-9)(4) = -36 and -9+4 = -5. Then we rewrite the -5x like -9x+4x and we factor like this:

3x^2 - 5x - 12=0 ----> 3x^2 - 9x + 4x - 12 =0 ---> 3x(x - 3) + 4(x - 3) = 0  -----> (x - 3)(3x + 4) = 0

x - 3 = 0 and  3x + 4 = 0 ---------> x = 3 and x = - 4/3

6x^2 + 9x - 6 = 0 --------->  6x^2/3 + 9x/3 - 6/3 = 0 ------>  2x^2 + 3x - 2 = 0  ------> 2x^2 + 4x - x - 2 = 0

2x(x + 2) - (x + 2) = 0  -------->  (x + 2)(2x - 1) = 0  --------> x + 2 = 0 and  2x - 1 = 0  ----> x = - 2   and x = 1/2

x^2 +3x - 28 = 0 -------> (x + 7)(x - 4) = 0 ---------> x + 7 = 0  and x - 4 = 0  -------> x = - 7  and  x = 4

3x^2 - 2x = 2x + 7  --------> 3x^2 - 2x - 2x - 7 = 0  -------> 3x^2 - 4x - 7 = 0 ------> 3x^2 + 3x - 7x - 7 = 0

3x(x + 1) - 7(x + 1) = 0  -------> (3x - 7)(x + 1) = 0  --------> 3x - 7 = 0  and  x + 1 = 0 -----> x = 7/3   and  x = -1

4x^2 -12x = 16  ----> (4x^2 - 12x - 16 = 0)/4  ------> x^2 - 3x - 4 = 0 ----->  (x - 4)(x + 1) = 0 ----> x = 4 and x = -1

x^2 + 3x = 0  -------> x(x + 3) = 0 -------> x = 0 and x = - 3

Please let me know if you have any doubt or question :)