It is difficult to understand the concept of competing the square so I will explain as far as possible:

The reason why"pramodpandey" removed `a` from `ax^2 +x-b` is because when completing the square, the co-efficient of `x^2` must be 1.

The co-efficient is the number in front of the variable (ie the `x` ).

To simplify his method a little, rearrange the equation:

`x^2 + x/a = b/a`

Now to complete the square you must halve the co-efficient of the `+x` which will then be squared. As the co-efficient is `1/a` = `(1/(2a))^2`

We are now going to use brackets and create a perfect square so we need to take `x^2` and bracket it to be `(x + ...)^2`

Note how the square now appears outside the bracket. Now add the halved (ie divided by 2) co-efficient `1/a` which you did before `= 1/(2a)` but do not square it now because it will be in the bracket which has the `^2` outside...ie `(x+1/(2a))^2`

Now do the right - hand side:

`therefore (x + 1/(2a))^2 = b/a + (1/(2a))^2`

It is an equation so you must always do to both sides and in terms of completing the square we add the `(1/2a)^2`

Simplify:

`(x+1/(2a))^2 = b/a + 1/(4a^2)`

Simplify further by using the LCD on the right hand side:

`(x+1/(2a))^2 = (4ab+1)/ (4a^2)`

To solve for x you need to get rid of the square so square root:

`x+1/(2a) = +- sqrt((4ab+1)/(4a^2))`

Note how a square root always renders a possible positive and negative answer ( `+-` ). The `sqrt(4a^2)` section simplifies to `2a` and we move the `+1/(2a)` to the other side so we get:

`x= -1/(2a) +- sqrt(4ab+1)/(2a)`

We can simplify as we have the same denominator

`x=( -1 +- sqrt(4ab+1))/(2a)` (as per pramodpandey)

`ax^2=b-x`

`ax^2+x-b=0`

`a(x^2+x/a-b/a)=0`

let a `!=0` ,then

`x^2+x/a-b/a=0` ,making perfect square

`x^2+2xx(x/(2a))+(1/(2a))^2-(1/(2a))^2-b/a=0`

`(x+1/(2a))^2-1/(4a^2)-b/a=0`

`` `(x+1/(2a))^2=(1+4ab)/(4a^2)`

Taking square root both side ,we have

`x+1/(2a)=+-sqrt((1+4ab)/(4a^2))`

`x=-1/(2a)+-sqrt(1+4ab)/(2a)`

`x=(-1+-sqrt(1+4ab))/(2a)`