1828 United States presidential election in Delaware
(Redirected from United States presidential election in Delaware, 1828)
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The 1828 United States presidential election in Delaware took place between October 31 and December 2, 1828, as part of the 1828 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for President and Vice President.
Delaware cast three electoral votes for the National Republican candidate, John Quincy Adams, over the Democratic candidate, Andrew Jackson. These electors were elected by the Delaware General Assembly, the state legislature, rather than by popular vote.[1]
Results
[edit]| 1828 United States presidential election in Delaware[2] | |||||
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| Party | Candidate | Votes | Percentage | Electoral votes | |
| National Republican | John Quincy Adams (incumbent) | – | – | 3 | |
| Democratic | Andrew Jackson | – | – | 0 | |
| Totals | – | – | 3 | ||
See also
[edit]References
[edit]- ^ "1828 Presidential General Election Results". U.S. Election Atlas. Retrieved April 13, 2013.
- ^ "Electoral Votes for President and Vice President 1821-1837". National Archives and Records Administration. Retrieved March 2, 2013.
State and district results of the 1828 United States presidential election | ||
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