United States presidential election in Georgia, 1828
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The 1828 United States presidential election in Alabama took place between October 31 and December 2, 1828, as part of the 1828 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for President and Vice President.
Georgia voted for the Democratic candidate, Andrew Jackson, over the National Republican candidate, John Quincy Adams. Jackson won Georgia by a margin of 93.58%.
Seven Georgian electors voted for William Smith for vice president, rather than Jackson's official running mate, John C. Calhoun.[1]
Results
United States presidential election in Georgia, 1828[2] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic | Andrew Jackson | 19,362 | 96.79% | 9 | |
National Republican | John Quincy Adams | 642 | 3.21% | 0 | |
Totals | 20,004 | 100.0% | 9 | ||
References
- ↑ "Electoral Votes for President and Vice President, 1821-1837 – 1828 Election for the Eleventh Term". National Archives and Records Administration. Retrieved 6 September 2018.
- ↑ "1828 Presidential General Election Results - Georgia". U.S. Election Atlas. Retrieved 28 February 2013.
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