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ladiesman217 Group Title

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  1. ladiesman217 Group Title
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    \[3x^2=33x+24\]

    • one year ago
  2. ladiesman217 Group Title
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    @ajprincess @nubeer @.Sam. can you help me please

    • one year ago
  3. ajprincess Group Title
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    \(3x^2=33x+24\) \(3x^2-33x+24=0\) Use quadratic formula. \[x=\frac{ -b\pm\sqrt{b^2-4ac} }{ 2a }\] the values of a, b and c can be found by comparing the equation \(3x^2-33x+24=0\) with the general equation \(ax^2+bx+c=0\). Does that help? @ladiesman217

    • one year ago
  4. ladiesman217 Group Title
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    ya

    • one year ago
  5. gohangoku58 Group Title
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    **-24 dividing the entire equation by 3 would simplify calculations

    • one year ago
  6. ajprincess Group Title
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    it is actually \(3x^2-33x-24=0\). am sorry abt the mistake

    • one year ago
  7. ajprincess Group Title
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    ya. first factor out 3 and then divide by 3 both sides. After that use quadratic formula:)

    • one year ago
  8. ladiesman217 Group Title
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    the equation i got was\[\frac{ -(-3)\sqrt{3^{2}-4(3)(-24)} }{ 2*3 }\]

    • one year ago
  9. ajprincess Group Title
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    b=-33

    • one year ago
  10. ladiesman217 Group Title
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    this is the correct equation right \[\frac{ -(-33)\sqrt{-33^{2}-4(3)(-24)} }{ 2*3 }\]

    • one year ago
  11. ajprincess Group Title
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    it will be \[\frac{ -(-33)\sqrt{(-33)^2-4*3*(-24)} }{ 2*3}\]

    • one year ago
  12. ladiesman217 Group Title
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    i got\[\frac{ 11\pm \sqrt{153} }{ 2 }\]

    • one year ago
  13. ajprincess Group Title
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    ya right:)

    • one year ago
  14. ajprincess Group Title
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    good work:)

    • one year ago
  15. ladiesman217 Group Title
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    thnx

    • one year ago
  16. ajprincess Group Title
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    welcome:)

    • one year ago
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