Deep breath....

TRIM(LOOKUP(99^99,--("0"&MID(A2,MIN(SEARCH({0,1,2,3,4,5,6,7,8,9},A2&"0123456789")),ROW(1:10000)))))

If you ask LOOKUP to look for a very large number (99 to the power of 99) in an array of numbers, it will return the last number in the array. This formula adds 0123456789 onto the end of the string in A2, so that EVERY digit is present at least once. The SEARCH bit then returns the POSITION in the string of the first occurrence of every digit in turn. The digit with the LOWEST value at this point HAS to be the first digit in the string. STARTING from that position (so that all the leading letters have been eliminated), it assembles a series of stings, being 1, 2, 3, 4, etc, characters long starting from that first digit. Sooner or later, it finds a space and thereafter text or a number, or whatever. From that point onwards, the value returned by the array ceases to be a number, its text consisting of a number and a space. Lookup returns the last real number it sees – which is the value that you want.

This: =TRIM(RIGHT(SUBSTITUTE(A2,"@",REPT(" ",LEN(A2))),LEN(A2))) returns everything after the last @ sign in a string. Put a space instead of the @ sign, and it will return 117 from A2.

Next is =SUBSTITUTE(A2, TRIM(RIGHT(SUBSTITUTE(A2," ",REPT(" ",LEN(A2))),LEN(A2))),””), which removes the 117 from the end of the string. Now all you need to do is repeat the earlier equation using this (and not A2) as the starting point.

=TRIM(RIGHT(SUBSTITUTE(SUBSTITUTE(A2, TRIM(RIGHT(SUBSTITUTE(A2," ",REPT(" ",LEN(A2))),LEN(A2))),””)," ",REPT(" ",LEN(SUBSTITUTE(A2, TRIM(RIGHT(SUBSTITUTE(A2," ",REPT(" ",LEN(A2))),LEN(A2))),””)))),LEN(SUBSTITUTE(A2, TRIM(RIGHT(SUBSTITUTE(A2," ",REPT(" ",LEN(A2))),LEN(A2))),””))))

Does just that. You now have the two numbers you need. All that remains is to put in the staring text and the odd hyphen.