1832 United States presidential election in Delaware

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United States presidential election in Delaware, 1832

← 1828 November 2 – December 5, 1832 1836 →
  Henry Clay.JPG Andrew Jackson.jpg
Nominee Henry Clay Andrew Jackson
Party National Republican Democratic
Home state Kentucky Tennessee
Running mate John Sergeant Martin Van Buren
Electoral vote 3 0
Popular vote 4,276 4,110
Percentage 50.99% 49.01%

The 1832 United States presidential election in Delaware took place between November 2 and December 5, 1832, as part of the 1832 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for President and Vice President.

Delaware voted for the National Republican candidate, Henry Clay, over the Democratic Party candidate, Andrew Jackson. Clay won Delaware by a margin of 0.98%.


United States presidential election in Delaware, 1832[1]
Party Candidate Votes Percentage Electoral votes
National Republican Henry Clay 4,276 50.99% 3
Democratic Andrew Jackson 4,110 49.01% 0
Totals 8,386 100.00% 3


  1. ^ "1832 Presidential General Election Results - Delaware". U.S. Election Atlas. Retrieved 12 April 2013.

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