United States presidential election in Delaware, 1936

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United States presidential election in Delaware, 1936

← 1932 November 3, 1936 1940 →
  Franklin D. Roosevelt - NARA - 196715.jpg LandonPortr.jpg
Nominee Franklin D. Roosevelt Alf Landon
Party Democratic Republican
Home state New York Kansas
Running mate John Nance Garner Frank Knox
Electoral vote 3 0
Popular vote 69,702 104,642
Percentage 54.62% 44.85%

President before election

Franklin D. Roosevelt
Democratic

Elected President

Franklin D. Roosevelt
Democratic

The 1936 United States presidential election in Delaware was held on November 3, 1936. The state voters chose three electors to the Electoral College, who voted for president and vice president.

Delaware voted for Democratic Party candidate and incumbent President Franklin D. Roosevelt, who defeated Republican nominee, Kansas Governor Alf Landon. Roosevelt won the state by a margin of 9.77 percent.

While Landon lost the state, the 44.85% of the popular vote made Delaware his fifth strongest state in the 1936 election in terms of popular vote percentage after Vermont, Maine, New Hampshire and Kansas[1].

Results

United States presidential election in Delaware, 1936[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Franklin Delano Roosevelt of New York John Nance Garner of Texas 69,702 54.62% 3 100.00%
Republican Alf Landon of Kansas Frank Knox of Texas 57,236 44.85% 0 0.00%
Union William Lemke of North Dakota Thomas C. O'Brien of Massachusetts 442 0.35% 0 0.00%
Socialist Norman Thomas of New York George A. Nelson of Wisconsin 172 0.13% 0 0.00%
Communist Earl Russell Browder of Kansas James W. Ford of New York 51 0.04% 0 0.00%
Total 127,603 100.00% 3 100.00%

References

  1. ^ "1936 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1936 Presidential General Election Results - Delaware". U.S. Election Atlas. Retrieved 23 December 2013.
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