Question 27815
Solve by completing the square.
x^2=5x+2
The golden rule for completing the square method:
Keep the square term and the term in x on the LHS.
When the  coefficient  of x^2 is 1 and the term in x is  positive, then
think of your perfect square in the form
[x+(1/2)times the coefficient of the term in x]^2 
Then expand using (a+b)^2 formula which gives  apart  from your two terms  on the left an additional quantity that is a constant.
And if coefficient  of x^2 is 1 and the term in x is  negative, then
think of your perfect square in the form
[x-(1/2)times the coefficient of the term in x]^2
Then expand using (a-b)^2 formula which gives  apart  from your two terms  on the left an additional quantity that is a constant. 
It is this constant that you have to add to both the  sides. You have a perfect square on theLHS and on the RHS you  have a sum of the given constant and the  constant you have  added.
Simplify the sum on the right.
Then take square root.
Do not forget to  give (+ or -)sign to the root on the right.
Transfer  the constant term from the left to the right.
You get two answers:
x= (the transfered constant+the rootwith the positive sign)
and x = (the transfered constant+the root with the negative sign))


x^2=5x+2
x^2-5x = 2  
We have (x-5/2)^2 = x^2-5x +25/4
Therefore we need 25/4 for the LHS to become a perfect square.
Adding 25/4 to both the sides,
x^2-5x +25/4= 2+25/4
That is (x-5/2)^2 = (8+25)/4
(x-5/2)^2 = (33)/4
Taking sqroot 
(x-5/2)=+ or -)sqroot(33/4)
x = [5/2+(or -)sqroot(33/4)]= [5/2+(or -)(1/2)sqroot(33)]

That is x = (5/2)+(1/2)[(33)^(1/2)]
And  x = (5/2)-(1/2)[(33)^(1/2)]
OR x = [5+ sqrt(33)]/2 and x = [5- sqrt(33)]/2