Complete the Square with $4x^2 - 4x + 3$ I am trying to complete the square of $4x^2 - 4x + 3$ and using the shortcut approach outlined here to get $d,e$ faster 
Here is my work so far, 
$$
\begin{align}
4x^2 - 4x + 3 &= 0  \\
x^2 - x + \frac{3}{4} &= 0 \\
x^2 - 2dx + (d^2 + e) &= 0 \\
d = - \frac{1}{2}, e = \frac{1}{2}
\end{align}
$$
However, this leads me to $(x - \frac{1}{2})^2 + \frac{1}{2}$. The answer in my textbook is $(2x - 1)^2 + 2$. I see how $x - \frac{1}{2} = 2x - 1$, but I don't get how the book gets 2 and why I got $\frac{1}{2}$ instead.
What did I do wrong? I feel like I am not using the first coefficient $a = 4$ correctly, but don't have a good reason as to why this is

edit:
So I guess I needed to add $a=4$ back to my equation and the correct form should have been $a(x+d)^2 + ae$, but I incorrectly learned it as $(x+d)^2 + e$!!!

You got a good result, just recall that you actually have
$$ax^2+bx+c=a(x-q)^2+ar$$
instead of $(x-q)^2+r$ for the final result.  Since $a=4$, you have
$$4\left(x-\frac 12\right)^2+4\cdot\frac 12\\
=(2x-1)^2+2.$$
