x²-2x-3 = 0 let's start by grouping the "x"s
(x² - 2x + [?])² - 3 = 0
so, we have a missing fellow there, in order to make the group, a perfect square trinomial, namely to "complete the square", hmmm so the tell-tale fellow is the middle term.
from a perfect square trinomial we know that the middle term is a product of 2 and the "term on the left" and the "term on the right", like
[tex]\bf \begin{array}{cccccllllll}
a^2& + &2 a b&+& b^2\\
\downarrow && &&\downarrow \\
a&& && b\\
&\to &( a + b)^2&\leftarrow
\end{array}\qquad \qquad \quad
\begin{array}{cccccllllll}
a^2& - &2 a b&+& b^2\\
\downarrow && &&\downarrow \\
a&& && b\\
&\to &( a - b)^2&\leftarrow
\end{array}[/tex]
[tex]\bf \textit{therefore}\qquad 2(x)(\boxed{?})=\stackrel{middle~term}{2x}\implies \boxed{?}=\cfrac{2x}{2x}\implies \boxed{?}=1[/tex]
aha!! so our missing fellow is 1.
now, let's keep in mind that all we're doing is borrowing from our very good friend Mr Zero, 0. So if we add 1², we also have to subtract 1².
(x² - 2x + 1² - 1²) - 3 = 0
(x² - 2x + 1) -1 -3 =0
(x - 1)² - 4 = 0
(x - 1)² = 4
