teach me how to complete the square when the x^2 value is higher than 1 please

STUDY THIS EXAMPLE SLOWLY AND CAREFULLY:

Solve for x:
4 x^2-8 x = 0

Divide both sides by 4:
x^2-2 x = 0

Add 1 to both sides:
x^2-2 x+1 = 1

Write the left hand side as a square:
(x-1)^2 = 1

Take the square root of both sides:
x-1 = 1 or x-1 = -1

Add 1 to both sides:
x = 2 or x-1 = -1

Add 1 to both sides:
Answer: |x = 2 or x = 0

Here's a pretty good tutorial on this topic :

http://www.purplemath.com/modules/sqrquad.htm

How about one like

3x^2 -10x +5 = 0   ?      Gets kinda messy, but it is done the same way.  DIvide thru by '3'

x^2 - 10/3 x = -5/3          Now take 1/2 of the 'x' term , square it and add to both sides

x^2 - 10/3 x  + 100/36  = -5/3  + 100/36      simplify

(x-10/6)^2  = 40/36     Take the sqrt of both sides

x- 10/6 = +- sqrt (40/36)

x =  10/6  +- sqrt40  / 6   = (10 +- sqrt40)/6