Can anybody please describe how and why it is even possible? If there is anything wrong what it is,?
Prove: $2+2=5$
$$-20=-20$$
$$16-36=25-45$$
$$16-36+\frac{81}{4}=25-45+\frac{81}{4}$$
$$4^2-2\cdot4\cdot\frac92+\left(\frac92\right)^2=5^2-2\cdot5\cdot\frac92+\left(\frac92\right)^2$$
$$\left(4-\frac92\right)^2=\left(5-\frac92\right)^2$$
$$4=5-\frac92+\frac92$$
$$4=5$$
$$2+2=5$$
On
Cancellation of the square is wrong in the 4th line from the bottom. $4-\frac{9}{2}=-\frac{1}{2}$ is negative, and $5-\frac{9}{2}=\frac{1}{2}$ is positive. They cannot be equal, but their squares are equal.
On
On the 4th from bottom line, you "cancel" the ${}^2$ sign. When doing this you should take the absolute value - $$\left|4-\dfrac92\right|=\left|5-\dfrac92\right|$$ which would be correct.
On
When they remove the powers is where they go wrong.
$$\bigg(4-\frac{9}{2}\bigg)^2 = \bigg(-\frac{1}{2}\bigg)^2 = \bigg(\frac{1}{2}\bigg)^2 = \bigg(5-\frac{9}{2}\bigg)^2$$
$$\bigg(-\frac{1}{2}\bigg) \neq\bigg(\frac{1}{2}\bigg)$$
On
\begin{align} & \left( 4 - \frac 9 2 \right)^2 = \left( 5 - \frac 9 2 \right)^2 \\[15pt] \text{Therefore } & 4 - \frac 9 2 = \pm\left( 5 - \frac 9 2 \right) \\[10pt] \text{i.e. } \quad & 4 - \frac 9 2 = \left\{\text{either } \quad +\left(5 - \frac 9 2\right) \quad\text{ or }\quad -\left( 5 - \frac 9 2 \right)\right\}. \end{align}
As the other answers have pointed out, the flaw lies in the step between $\left(4-\frac92\right)^2=\left(5-\frac92\right)^2$ and $\left(4-\frac92\right)=\left(5-\frac92\right)$. There are a number of ways to explain why you can't do this:
One reason you can't do it is because $\sqrt{x}$ takes $2$ different values. Namely, $\sqrt{x}$ and $-\sqrt{x}$. This is because $x^2=(-x)^2$. When you do the incorrect step in your proof, you're taking the square root of both sides but really, you should be taking either $\sqrt{x}$ or $-\sqrt{x}$. So, in fact, it would be correct to say that $\left(4-\frac92\right)=-\sqrt{\left(5-\frac92\right)^2}=-\left(5-\frac92\right)$.
Another way of explaining why that step is wrong is that for any number, $x$, you get the same number from $x^2$ as $(-x)^2$. Case in point, $3^2=(-3)^2$ and, as isyoung points out, $0.5^2=(-0.5)^2$. So, this fake-proof is designed to have one side equal to $0.5^2$ but the other side equal to $(-0.5)^2$. Specifically, the fake-proof says that $(4-4.5)^2=(5-4.5)^2$ since $4-4.5=-0.5$ and $5-4.5=0.5$. This trick lets the fake-proof have you thinking that $0.5=-0.5$. All other parts of the fake-proof are just puffery disguising that step.
In summary, $x^2=y$ doesn't just mean $x=\sqrt{y}$, it can also mean $x=-\sqrt{y}$.