Completing the Square?

Question

i know the answer but i need see these problems work out

x2-4x+3=0 and x2+3x-5=0

Answer 1

x² - 4x + 3 = 0

x² - 4x + 3 - 3 = 0 - 3

x² - 4x = - 3

x² - 4 +_______= - 3 +______

x² - 4x + 4 = - 3 + 4

x² - 4x + 4 = 1

(x - 2)(x - 2) = 1

(x - 2)² = 1

(√x - 2)² = ± √1

x - 2 = ± 1

x - 2 + 2 = 2 ± 1

x = 2 ± 1

- - - - - - - - -

Solving for +

x = 2 + 1

x = 3

- - - - - - -

Solving of -

x = 2 - 1

x = 1

- - - - - - - s-

Answer 2

x^2 - 4x + 3 = 0

1. move the 3 on over to the other side
x^2 - 4x = -3

2. take 1/2 of the term with the x (-4)
so will be -2

3. now square the anwser and add it to both sides
-2^2 = 4
x^2 - 4x + 4 = -3 + 4
x^2 - 4x + 4 = 1

4. now you can write the right side as a square
(x-2)^2 = 1

5. take the square root of both sides
sq root of (x-2)^2 = x-2
sq root of 1 = 1

6. so have x-2 = 1

7 solve for x
x - 2 = 1
x = 3


Same for the next one3

x^2 + 3x - 5 = 0
x^2 + 3x = 5

1/2 of 3 = 3/2
3/2^2 = 9/4

x^2 + 3x + 9/4 =29/4
(x + 3/2)^2 = 29/4
x + 3/2 = sq rt 29/4
x = (sq rt 29 - 3)/2

Answer 3

x² - 4x + 3 = 0
(x - 2)² - 1 = 0
(x - 2)² = 1
x-2 = +/- √1
x = 2 +/- √1


x² + 3x - 5 = 0
(x + 3/2)² - 29/4 = 0
(x + 3/2) = √(29/4)
x = -3/2 +/- (√29)/2

Answer 4

x^2-4x+4-4+3=0
(x-2)^2-1=0
(x-2+1)(x-2-1)=0
(x-1)(x-3)=0

2.x^2+3x+9/4-9/4-5=0
(x+3/2)^2-29/4=0
(x+3/2)-{(rt29)/2}^2=0
(x+3/2+rt29/2)(x+3/2-rt29/2)=0
[x+(3+rt29)/2][x+(3-rt29)/2)]=0