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?