Why is $x=2 \implies (x-2)(x-3)=0$ false? Let $P(x)$ be the equation $x=2$
and $Q(x)$ be the equation  $(x-2)(x-3)=0$

By definition of implication I see that $P(x)$ implies $Q(x)$...

As I see it, any premise that is false can give any consequence.
With $x=2$ both sides of the arrow are true.. so implication is true.

It is clear that $Q(x)$ implies $P(x)$ and it is also clear that $P(x)$ and $Q(x)$ are not equivalent. Since $Q(x)$ implies $P(x)$ and $P(x)$ and $Q(x)$ are not equivalent it follows that $P(x)$ cannot imply $P(x)$.

Where am I wrong?

EDIT: I knew that $P(x)\Leftrightarrow Q(x)$ was not valid. And I thought wrongly that $P(x)\Leftarrow Q(x)$ is valid which should, if it was valid, imply that $P(x)\nRightarrow Q(x)$. But testing $P(x)\Rightarrow Q(x)$ by the truth table showed $P(x)\Rightarrow Q(x)$ is valid which confused me. I was simply wrongly thinking $P(x)\Leftarrow Q(x)$ which is the correct answer. I have also posted a soultion below, which typo is corrected.
$$(x-2)(x-3)=0 \iff x = 2 \ \, \text{or} \, \ x = 3$$

If $x=2$, then the hypothesis of the RHS is satisfied and hence we have the LHS.

I don't see why that is supposed to be false.
For the case $P(x)\Rightarrow Q(x)$:

If we choose $x=2$ then we got true on both sides of the arrow. If we choose $x\neq 2$ then P(x) is false and Q(x) can be false (if $x\neq 3$) or true (if $x=3$).
The implication $P(x)\Rightarrow Q(x)$ is therefore valid.

Following explanation will explain why I was wrong. This is for the case $P(x)\Leftarrow Q(x)$:

If we choose $x=2$ then we got true on both sides of the arrow.
If we choose $x=3$ then we got Q(x) is true and P(x) is false. Since this scenario is possible, it means the implication $P(x)\Leftarrow Q(x)$ is not valid, i.e. we have $P(x)\nLeftarrow Q(x)$.

A bit more intuitive and more self explaining similar example is when we consider $x=3 \Rightarrow x^2=9$ which is fully valid. What if we change the direction of the arrow in the implication? Changing direction of the arrow, gives us $x=3 \nLeftarrow x^2=9$ since $x=-3$ is true for $x^2=9$ but not for $x=3$.