Solving a quadratic equation by completing the square method

I will demonstrate by two examples. Example 1 has the coefficient of x^2=1 and Example 2 has the coefficient of x^2 greater than 1
Example 1
solve by completing the square method; x^2-7x+10=0 (note also the coefficient of x =-7)
solution
The first step is to take the constant 10 to the RHS to get;
x^2-7x=-10
Next we add a constant k to both sides to get;
x^2-7x+k=-10+k ....(*)
Next we find the value of k to 'complete the square' on the LHS by making it a perfect square i.e.;
k=(1/2*the coefficient of x)^2
k=(1/2*-7)^2=(-7/2)^2
Now we write equation (*) as follows;
x^2-7x+(-7/2)^2=-10+(-7/2)^2
The LHS can be easily factorized as below;
(x-7/2)^2=-10+49/4
(x-7/2)^2=9/4
We next find the square root of both sides to get;
x-7/2=±3/2
x=7/2±3/2
x=3.5±1.5
x=5 or 2
Example 2
Solve by completing the square method; 2x^2+x-3=0 (Note that the coefficient of x^2=2 unlike example 1 above)
Our first step is to make the coefficient of x^2=1 by dividing every term by 2 to get;
x^2+x/2-3/2=0
Next we take the constant to the RHS to get;
x^2+x/2=3/2
Next we add a k to both sides to get;
x^2+x/2+k=3/2+k.....(**)
Next we find the value of k to 'complete the square' on the LHS by making it a perfect square i.e.;
k=(1/2*coefficient of x)^2=(1/2*1/2)^2=(1/4)^2
Equation (**) can now be written as;
x^2+x/2+(1/4)^2=3/2+(1/4)^2
the LHS can be factorized to get;
(x+1/4)^2=3/2+1/16
(x+1/4)^2=9/16
Next find the square of both sides to get;
x+1/4=±3/4
x=/1/4±3/4= 4/4 or -2/4 
x=1 or -1/2