SOLUTION: solve for x: 80x^2-96x+27=0

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Question 1069121: solve for x: 80x^2-96x+27=0
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
80x^2-96x+27=0
x=(1/160)(96+/- sqrt (96^2-8640); sqrt (9216-8640)=sqrt (576)=24
x=(1/160)(96+/-24)
x=(1/160)(120), or 3/4
x=(1/160)(72) or 9/20
80(9/16)-96(3/4)+27=45-72+27=0
80(.2025)-96(.45)+27=16.2-43.2+27=0
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