HELP ASAP PLEASE

The squre root has to be a 4 or 6  for the square to end in 6

4         16

6             36

14          196

16        256

24     576

 26       676

34            1156

36           1296

44        1936

46        2116                             So I'd say    1  3  9   5  or   7    for the tens' digit.....

squares that end in 6 must be of the form

$(10n\pm4)^2$     where n is an interger 

$(10n\pm4)^2\\ =100n^2\pm80n+16\\ =100n^2+10\pm80n+6\\ =100n^2+10(1\pm8n)+6\\ \text{8n can end in any even number so }1\pm8n \text { can end in any odd number}\\ \text{So the 10s place value can have any odd digit.}$