CREATE TABLE table_21436373_7 (date_year VARCHAR2, result_games VARCHAR2)

When 9 games (28,002 avg.) is the results of the games what is the date/year?
SELECT date_year FROM table_21436373_7 WHERE result_games = "9 games (28,002 avg.)"
SELECT "date_year" FROM "table_21436373_7" WHERE "9 games (28,002 avg.)" = "result_games"