<s>[INST] 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, ($a x^2$)?)
 [/INST] What you have done is correct so far. Nice job! Next time, you could divide both sides of $2x^2-9x+9 = 0$ to get $x^2-\frac{9}{2}x + \frac{9}{2}=0$.
From your method, we have:
$$(x-\frac{9}{4})^2 - (\frac{9}{4})^2+\frac{9}{2} =0$$
$$(x-\frac{9}{4})^2 = \frac{81}{16} - \frac{9}{2}$$
$$(x - \frac{9}{4})^2 = \frac{9}{16}.$$
The next step is very important. When taking the square root, there are always two solutions — one positive and one negative.
$$x - \frac{9}{4} = \color{red}{±} \color{black}{\frac{3}{4}}$$
From here you can find the two values of $x$, and then factorise.
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