Completing the square

This problem is bothering me ..
Completr the square
4x^2-4x+3
pleaseeeee

alright so...
1) 4x^2-4x+3 .... first u have to common factor leaving the 3 out of it
----> 4(x^2-x) +3
2) since x is 1.. u take it and add 1, divide by 2 and then square it.. then you also have to subtract 1, divide by 2, and square it
-----> 4[x^2 - x +(1/2)^2 - (1/2)^2] +3
3) next you take (x^2-x+(1/2)^2) and you factor it
-----> 4[(x-1/2)^2 - (1/2)^2] +3
4) next you take the 1/2^2 and solve which you get 1/4
----->4[(x-1/2)^2 -1/4] +3
5) next we have to take the -1/4 out of the bracket so we times it by 4
----> 4(x-1/2)^2 -1+3
6) next just solve -1+3 which you get 2
-----> 4(x-1/2)^2 +2
and thats it .. to sum it up this is all the steps
4x^2-4x+3
4(x^2-x) +3
4[x^2 - x +(1/2)^2 - (1/2)^2] +3
4[(x-1/2)^2 - (1/2)^2] +3
4[(x-1/2)^2 -1/4] +3
4(x-1/2)^2 -1+3
4(x-1/2)^2 +2

4x² - 4x + 3
Isolate the variables onto one side, and the constants on the other
4x² - 4x = -3
Divide the whole equation by 4

x² - 1 = -3 / 4
Now, x² - 1 = (x + 1)(x - 1) = - 3/4

As you can see, you cannot complete the square. This can be shown if you take the discriminant of the expression, which is negative. Therefore, this expression can not be applied to the quadratic formula I hope this helps.

http://www.youtube.com/watch?v=xGOQYTo9AKY

http://youtu.be/zKV5ZqYIAMQ

!! the answer to your equation is an imaginary number :)

4x^2 -- 4x + 3
= [4x^2 -- 4x + 1] + 2
= (2x -- 1)^2 -- [sqrt(2) i]^2
= (2x -- 1 + sqrt (2) i )(2x -- 1 -- sqrt(2) i)

4x² - 4x + 3
(2x - 2)² - 1