We can solve the given equation using the two methods:

For the first,

x^2 - 4x + 3 = 0

express -4 as the sum of two numbers which give the product 3

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

=> x(x - 3) -1(x -3) =0

=> (x-1)(x-3) =0

(x -1) =0 => x= 1

and (x-3)= 0 => x = 3

So we get the roots 1 and 3

Also we can use the formula for calculating roots directly

x= [ -b + sqrt[ b^2- 4ac]/ 2a and x= [ -b - sqrt[ b^2- 4ac]/ 2a

Here a= 1, b = -4 and c=3

x= [ -b + sqrt[ b^2- 4ac]/ 2a

= [ 4 + sqrt{(-4)^2- 4*3}]/ 2

= [ 4 + sqrt 4 ]/ 2

= [ 4 + 2 ]/2

= 3

x= [ -b - sqrt[ b^2- 4ac]/ 2a

= [ 4 - sqrt{(-4)^2- 4*3}]/ 2

= [ 4 - sqrt 4 ]/ 2

= [ 4 - 2 ]/2

= 1

Again the roots are 1 and 3

**So the roots we get are 1 and 3.**

To find the roots in 2 methods: x^2-4x+3 = 0

Let x1 and x2 be the roots.

x^2-4x+3 = k(x-x1)(x-x2) are identically equal by remainder theorem.

When k = 1, both sides have the same leading terms x^2 with 1 as coefficients. So we equate the coefficients of like powers of x and constant terms on both sides of:-

x^2-4x+3 = x^2-(x1+x2)x +x1x2.

Coefficient of x:

-4 = -( x1+x2). Or

x1+x2 = 4.........(1)

x1x2 = 3

Therefore x1-x2 = sqrt{ (x1+x2)^2 -4x1x2)} = sqrt(4^2 -4*3 )= sqrt4 = 2.

x1-x2 = 2 ..........(2)

x1+x2 = 4.........(1)

Adding the equations (1) and (2): 2x1 = 6, x 1 = 3

Subtracting eq(2) from eq(1), we get: 2x2 = 4-2 = 2. So, x2 = 1

x1 = 3 and x2 = 1.

2nd method:

x^2 -4x+3 = 0

x^2-4x = -3

Add 2^2 =4 to both sides so that LHS is a perfect square of (x-2)^2.

x^2-4x+ 4 = 4-3

(x-2)^2 = 1

Take square root:

x-2 = 1 or x-2 = -1

x = 2+1 = 3.

x = 2-1 = 1.

**The first method is to use factorization:**

The original equation is: x^2-4x+3=0

We'll write 3 as the difference 4 - 1 = 3

We'll re-write the equation, changing 3 with the difference:

x^2-4x+4-1=0

Now, we'll combine the first and the last terms and the middle terms:

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

We'll write the difference of squares as a product:

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

We'll factorize (4x - 4) by 4:

(4x - 4) = 4(x-1)

Now, we'll re-write the equation:

(x-1)(x+1) - 4(x-1) = 0

We'll factorize by (x-1) and we'll get:

(x-1)(x+1-4) = 0

We'll set each factor as zero:

x - 1 = 0

We'll add 1 both sideS:

x = 1

x - 3 = 0

We'll add 3 both sides:

x = 3

**The roots of the equation are: {1 ; 3}. **

**The second method is to apply the quadratic formula:**

x1 = [-b+sqrt(b^2-4ac)]/2a

x1 = [4+sqrt(16-12)]/2

x1 = (4+2)/2

**x1 = 3**

x2 = (4-2)/2

**x2 = 1**

**As we can see, we've get the same result!**