Election in Delaware
1824 United States presidential election in Delaware
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← 1820 | October 26 – December 2, 1824 | 1828 → |
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| | | Nominee | William H. Crawford | John Quincy Adams | | Party | Democratic-Republican | Democratic-Republican | Home state | Georgia | Massachusetts | Running mate | Nathaniel Macon | John C. Calhoun | Electoral vote | 2 | 1 | |
President before election James Monroe Democratic-Republican | Elected President John Quincy Adams Democratic-Republican | |
Elections in Delaware |
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The 1824 United States presidential election in Delaware took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for President and Vice President.
During this election, the Democratic-Republican Party was the only major national party, and four different candidates from this party sought the Presidency. Delaware cast two electoral votes for William H. Crawford and one for John Quincy Adams.
Results
See also
References
- ^ "Electoral Votes for President and Vice President 1821-1837". National Archives and Records Administration. Retrieved February 28, 2013.