1896 United States presidential election in Delaware
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![]() County Results
McKinley 50-60%
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Elections in Delaware |
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The 1896 United States presidential election in Delaware took place on November 3, 1896. All contemporary 45 states were part of the 1896 United States presidential election. State voters chose three electors to the Electoral College, which selected the president and vice president.
Delaware was won by the Republican nominees, former Ohio Governor William McKinley and his running mate Garret Hobart of New Jersey. They defeated the Democratic nominee, former U.S. Representative from Nebraska William Jennings Bryan and his running mate Arthur Sewall. McKinley won the state by a margin of 10.08%.
As a result of his win in Delaware, McKinley became the first Republican presidential candidate since Ulysses S. Grant in 1872 to win the state.
Bryan would lose Delaware to McKinley again four years later and would later lose the state again in 1908 to William Howard Taft.
Results
[edit]1896 United States presidential election in Delaware[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Republican | William McKinley | 20,450 | 53.18% | 3 | |
Democratic | William Jennings Bryan | 16,574 | 43.10% | 0 | |
National Democratic | John M. Palmer | 966 | 2.51% | 0 | |
Prohibition | Joshua Levering | 466 | 1.21% | 0 | |
Totals | 38,456 | 100.00% | 3 | ||
Voter turnout | — |
See also
[edit]Notes
[edit]References
[edit]- ^ Dave Leip's U.S. Election Atlas; Presidential General Election Results – Delaware