Basically, this part of the function finds the location of the second-to-last space in the string:

FIND("@",SUBSTITUTE(A2," ","@",LEN(A2)-LEN(SUBSTITUTE(A2," ",""))-1))

The "@" is used as a temporary placeholder and it is inserted via the Substitute function in the place of interest.

The following part of the function calculates the number of spaces in the string: LEN(A2)-LEN(SUBSTITUTE(A2," ","") by taking the length of the original string and subtracting the length of the string after all spaces have been substituted with nulls.

The main Substitute function then replaces the second-to-last space (determined by subtracting 1 from the total number of spaces found with the previously described substitute function) with an "@".

Then the FIND() function locates the position of that "@" and uses that to determine how many characters to take in the main LEFT() function.

Actually, come to think of it, I did neglect to subtract another 1 character. The formula I gave you actually retains the space at the end...use the following revision instead.

=IF(ISNUMBER(VALUE(LEFT(RIGHT(A2,3),1))),LEFT(A2,FIND("@",SUBSTITUTE(A2," ","@",LEN(A2)-LEN(SUBSTITUTE(A2," ",""))-1))**-1**),A2)