1992 United States presidential election in Georgia

1992 United States presidential election in Georgia

← 1988 November 3, 1992 1996 →
Turnout47%
 
Nominee Bill Clinton George H. W. Bush Ross Perot
Party Democratic Republican Independent
Home state Arkansas Texas Texas
Running mate Al Gore Dan Quayle James Stockdale
Electoral vote 13 0 0
Popular vote 1,008,966 995,252 309,657
Percentage 43.47% 42.88% 13.34%

County Results

President before election

George H. W. Bush
Republican

Elected President

Bill Clinton
Democratic

The 1992 United States presidential election in Georgia took place on November 3, 1992, and was part of the 1992 United States presidential election. Voters chose 13 representatives, or electors to the Electoral College, who voted for president and vice president.

Georgia was won by Governor Bill Clinton (D-AR). The presidential contest in Georgia was the closest of any state that year, with Clinton winning 43.47% to 42.88% over Incumbent President George H. W. Bush (R-TX) by a thin margin of 0.59%. This made it the first time that Georgia had voted Democratic since 1980, when it voted for Jimmy Carter, who was the former Governor. The state flipped back into the Republican column in 1996 and stayed there until 2020.[1] From the mid-1960s into the 1990s, Georgia was a swing state in presidential elections, but also a state where Democrats generally dominated at the state and local level. Billionaire businessman Ross Perot (I-TX) finished in third, with a significant 13.34% of the popular vote.[2]

  1. ^ Cite error: The named reference :0 was invoked but never defined (see the help page).
  2. ^ "1992 Presidential Election results". sos.Georgia.gov.

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