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rebecca1233

  • 3 years ago

Solve by using the quadratic formula. 3x^2 - 4x = 2

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  1. rebecca1233
    • 3 years ago
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    @jasonxx

  2. brandonloves
    • 3 years ago
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    I could be totally off, but I just wanted to give it a shot. I got 8/3, 3 again, I could be way wrong, but I just wanted to try it/see if I remembered. good luck you know the formula right?

  3. rebecca1233
    • 3 years ago
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    |dw:1352320744523:dw|

  4. rebecca1233
    • 3 years ago
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    |dw:1352320782461:dw|

  5. rebecca1233
    • 3 years ago
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    |dw:1352320879213:dw|

  6. rebecca1233
    • 3 years ago
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    or none of the above

  7. brandonloves
    • 3 years ago
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    |dw:1352320868071:dw| I think I'm right up till this point, but I think I get messed up after that.

  8. rebecca1233
    • 3 years ago
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    the answer is the last one i believe?

  9. brandonloves
    • 3 years ago
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    oh, I see how you got the 40. isn't it -4(a)(c) though? I got -24 + 16 to get -8 under the radical

  10. rebecca1233
    • 3 years ago
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    |dw:1352321326332:dw|

  11. rebecca1233
    • 3 years ago
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    idk im confused

  12. brandonloves
    • 3 years ago
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    http://www.purplemath.com/modules/quadform.htm

  13. rebecca1233
    • 3 years ago
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    @.UserNotFound. am i correct because

  14. rebecca1233
    • 3 years ago
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    i need help

  15. .UserNotFound.
    • 3 years ago
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    http://answers.yahoo.com/question/index?qid=20091117132621AAECMYZ ??

  16. brandonloves
    • 3 years ago
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    yea, and I'm not much help haha. I tried though. I'd check out khanacademy.org or watch some youtube videos till someone with more quadform exp can help hah

  17. rebecca1233
    • 3 years ago
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    i just wanna know if i am correct

  18. jasonxx
    • 3 years ago
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    let me check

  19. jasonxx
    • 3 years ago
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    \[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }=\frac{ 4 \pm \sqrt{16-4*3*(-2)} }{ 2*3 }= \frac{ 4 \pm \sqrt{40} }{ 6 }\] @rebecca1233

  20. jasonxx
    • 3 years ago
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    @rebecca1233 come on now show me , yor answer is here

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