mathematics - Something is wrong with the equation

There is something wrong with the equation below:

By just exchanging two squares at a time, find the equality with the least amount of exchange.

For example if this question is asked for below equation, you may find the solution with one time exchange:

the answer will be exchanging "+" and "9" which makes the equation correct as below:

$29+5=34$

FYI: Even though it is not a hint, all numbers are used only once. ($0,1,2,3,4,5,6,7,8,9$)

Hint: you only swap each square once for optimal clear solution.

Answer

@oray's edit: Here is my intented answer since no-one found it out:

$3140\times(6+9)=2-75\times8$ (Initial layout)
$3140\times(6+9)-2=75\times8$ ("$=$" and "$-$")
$3\times40\times(6+9)-2=7518$ ("$\times$" and "$1$")
$3\times45\times(6+9)-2=7018$ ("$0$" and "$5$")

$3\times45\times(6+9)-7=2018$ ("$2$" and "$7$")

which is

the year we are in.

I found another solution with

4 exchanges

which is:

$3140\times(6+9)=2-75\times8$ (Initial layout)
$3+40\times(619)=2-75\times8$ (Swapped $1$ and $+$)
$3+40=(619)\times2-75\times8$ (Swapped $=$ and $\times$)
$3+40=(629)\times1-75\times8$ (Swapped $2$ and $1$)
$5+40=(629)\times1-73\times8$ (Swapped $5$ and $3$)
To arrive at $45=45$

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Original post: I'm not sure if this is optimal, but I can do it in

4 exchanges

like this:

$3140\times(6+9)=2-75\times8$ (Initial layout)
$314\times0(6+9)=2-75\times8$ (Swapped 0 and $\times$)
$314\times0(6+9)=27-5\times8$ (Swapped 7 and $-$)
$314\times0(6+8)=27-5\times9$ (Swapped 8 and 9)

$514\times0(6+8)=27-3\times9$ (Swapped 3 and 5)
To arrive at $0=0$.