algebra precalculus - What is going wrong in this "proof" of $0=1$?

\begin{align}
-20 &= -20\\
16-36 &= 25-45\\
4^2-4\times 9&=5^2-5\times 9\\
4^2-4\times 9+81/4&=5^2-5\times 9+81/4\\
4^2-4\times 9+(9/2)^2&=5^2-5\times 9+(9/2)^2\\

\end{align}

Considering the formula $a^2+2ab+b^2=(a-b)^2$, one has
\begin{align}
(4-9/2)^2&=(5-9/2)^2\\
\sqrt{(4-9/2)^2}&=\sqrt{(5-9/2)^2}\\
4-9/2&=5-9/2\\
4&=5\\
4-4&=5-4\\
0&=1

\end{align}

Answer

Here let me simplify your process:

$$
\begin{align}
(-2)^2 = 4 &\implies \sqrt{(-2)^2} = \sqrt{2^2} \\
&\implies -2 = 2 \\
&\implies -2 + 2 = 2 +2 \\
&\implies 0 = 4

\end{align}
$$

QED. Do you see the mistake?