How to factorize $2x^2-9x+9$ by completing the square? I know that $x^2-bx+c=(x-k)^2=x^2-2kx+k^2$ if it is a complete square. If not we create one by adding and subtracting $\left(\frac{b}{2}\right)^2$
I tried $$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x^2-\frac{9}{2}x+\left(\frac{9/2}{2}\right)^2-\left(\frac{9/2}{2}\right)^2+\frac{9}{2}\right)$$
What am I missing and what is the best way to factorize when we have coefficients?(Is this the right term, ($a x^2$)?)

$$2x^2-9x+9\tag1$$
Factorise 
$$2\left(x^2-{9\over 2}x+{9\over 2}\right)\tag2$$
$$2\left(x^2-{9\over 2}x+{9\over 2}\right)=2\left[\left(x-{9\over 4}\right)^2+{9\over 2}-\left(9\over 4\right)^2\right]\tag3$$
$$2\left(x^2-{9\over 2}x+{9\over 2}\right)=2\left[\left(x-{9\over 4}\right)^2-{9\over 16}\right]\tag4$$
$$2x^2-9x+9=2\left(x-{9\over 4}\right)^2-{9\over 8}\tag5$$
