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Posted by on Tuesday, July 26, 2011 at 12:42pm.

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/b2-4ac; b=3, a=-1, c=-2
What would the solution be?

  • algebra - , Tuesday, July 26, 2011 at 12:54pm

    b^2-4ac=9-8=1
    X1=(-b-sqrt(1))/2a=(-3-1)/(-2)=2
    X2=(-b+sqrt(1))/2a=(-3+1)/(-2)=1

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