calculusxy
  • calculusxy
MEDAL!!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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calculusxy
  • calculusxy
\[\huge 3n^2 - 11 = 8n \]
AlexandervonHumboldt2
  • AlexandervonHumboldt2
solve for n?
calculusxy
  • calculusxy
yes using quadratic formula

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calculusxy
  • calculusxy
\[\large 3n^2 - 8n - 11 = 0\] would that equation be correct?
anonymous
  • anonymous
yes
AlexandervonHumboldt2
  • AlexandervonHumboldt2
use quadratic formula \[\frac{ -b \pm \sqrt{b^2-4ac }}{ 2a }\] where ax^2+bx+c=0 yes this equation is correct substitute values:
calculusxy
  • calculusxy
\[n = \frac{ 8 \pm \sqrt{64 - 4(-33)} }{ 6 }\]
anonymous
  • anonymous
looks good
AlexandervonHumboldt2
  • AlexandervonHumboldt2
yeah correct now simplify this
calculusxy
  • calculusxy
\[n = \frac{ 8 \pm \sqrt{196} }{ 8 } = \frac{ 8 + 14 }{ 8 }\]
AlexandervonHumboldt2
  • AlexandervonHumboldt2
how denominator became 8 from 6?
anonymous
  • anonymous
how did your 6 become an 8 in the denominator
AlexandervonHumboldt2
  • AlexandervonHumboldt2
everything is correct exept the denominator
calculusxy
  • calculusxy
sorry typo :\
AlexandervonHumboldt2
  • AlexandervonHumboldt2
ok
anonymous
  • anonymous
ah ;)
AlexandervonHumboldt2
  • AlexandervonHumboldt2
it would be \[\frac{ 8 \pm 14 }{ ? }\]
AlexandervonHumboldt2
  • AlexandervonHumboldt2
woops 6 instead of ?
calculusxy
  • calculusxy
(3.66, -1)
AlexandervonHumboldt2
  • AlexandervonHumboldt2
correct
calculusxy
  • calculusxy
thank you @Lily2913 @AlexandervonHumboldt2
AlexandervonHumboldt2
  • AlexandervonHumboldt2
np :)
anonymous
  • anonymous
np :)

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