Question 304623
find the real solutions, using the quadratic equation 
3x(x+2)=1 

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3x^2+6x-1 = 0
1x = [-6 +- sqrt(36-4*3*-1)]/6
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x = [-6 +- sqrt(48)]/6
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x = [-6 +-4sqrt(3)]/6
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x = -1+(2/3)sqrt(3) or x = -1-(2/3)sqrt(3)
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Cheers,
Stan H.