1988 United States presidential election in Delaware

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United States presidential election in Delaware, 1988

← 1984 November 8, 1988 1992 →
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Nominee George H. W. Bush Michael Dukakis
Party Republican Democratic
Home state Texas Massachusetts
Running mate Dan Quayle Lloyd Bentsen
Electoral vote 3 0
Popular vote 139,639 108,647
Percentage 55.88% 43.48%

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County Results

Bush

  50–60%
  60–70%

President before election

Ronald Reagan
Republican

Elected President

George H. W. Bush
Republican

The 1988 United States presidential election in Delaware took place on November 8, 1988. All 50 states and the District of Columbia, were part of the 1988 United States presidential election. Delaware voters chose 3 electors to the Electoral College, which selected the president and vice president.

Delaware was won by incumbent United States Vice President George H. W. Bush of Texas, who was running against Massachusetts Governor Michael Dukakis. Bush ran with Indiana Senator Dan Quayle as Vice President, and Dukakis ran with Texas Senator Lloyd Bentsen.

Delaware weighed in for this election as about 4% more Republican than the national average. As of the 2016 presidential election, this is the last time that the Republican nominee carried Delaware.

Partisan background

The presidential election of 1988 was a very partisan election for Delaware, with nearly 99.5% of the electorate voting for either the Democratic or Republican parties, and only four political parties on the ballot, statewide.[1] All counties in Delaware voted in majority for Bush. As of the 2016 presidential election, this is the last election in which New Castle County voted for the Republican candidate.

Results

United States presidential election in Delaware, 1988
Party Candidate Votes Percentage Electoral votes
Republican George H. W. Bush 139,639 55.88% 3
Democratic Michael Dukakis 108,647 43.48% 0
Libertarian Ron Paul 1,162 0.47% 0
New Alliance Party Lenora Fulani 443 0.18% 0
Totals 249,891 100.00% 3

See also

References

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