ladiesman217
  • ladiesman217
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Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ladiesman217
  • ladiesman217
\[3x^2=33x+24\]
ladiesman217
  • ladiesman217
@ajprincess @nubeer @.Sam. can you help me please
ajprincess
  • ajprincess
\(3x^2=33x+24\) \(3x^2-33x+24=0\) 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

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

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