ladiesman217
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ladiesman217
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\[3x^2=33x+24\]
ladiesman217
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@ajprincess @nubeer @.Sam. can you help me please
ajprincess
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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
ladiesman217
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ya
gohangoku58
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**-24
dividing the entire equation by 3 would simplify calculations
ajprincess
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it is actually \(3x^2-33x-24=0\). am sorry abt the mistake
ajprincess
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ya. first factor out 3 and then divide by 3 both sides. After that use quadratic formula:)
ladiesman217
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the equation i got was\[\frac{ -(-3)\sqrt{3^{2}-4(3)(-24)} }{ 2*3 }\]
ajprincess
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b=-33
ladiesman217
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this is the correct equation right
\[\frac{ -(-33)\sqrt{-33^{2}-4(3)(-24)} }{ 2*3 }\]
ajprincess
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it will be
\[\frac{ -(-33)\sqrt{(-33)^2-4*3*(-24)} }{ 2*3}\]
ladiesman217
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i got\[\frac{ 11\pm \sqrt{153} }{ 2 }\]
ajprincess
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ya right:)
ajprincess
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good work:)
ladiesman217
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thnx
ajprincess
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welcome:)