paying algebra help equations!!

Discussion in 'Archives' started by kamaka, Dec 3, 2009.

paying algebra help equations!!
  1. Unread #1 - Dec 3, 2009 at 12:37 AM
  2. kamaka
    Joined:
    Aug 3, 2009
    Posts:
    44
    Referrals:
    0
    Sythe Gold:
    0

    kamaka Member
    Banned

    paying algebra help equations!!

    I'm pretty much dumb when it comes to math....anyways i need help with these...I can pay whoever can help me finish these by paypal I have a canadian account I can offer $4-5

    Find the discriminant

    3x^2+6x-8=0
    I already know the a,b,c

    a=3,b=6,c=-8

    5x^2+3x=12 (have to put in standard form first
    7-5x^2+9x=x (have to put in standard form first i think
    2x=x^2-x (have to put in standard form first

    Determine the number of solutions in the equation one, two, or none

    x^2-3x+2=0
    -3x^2+5x-1=0
    x^2-2x+4=0
    3x^2-6x+3=0
    -5x^2+6x-6=0

    Thanks please private message me or you can put them as a response and i can pay you then thanks:)
     
  3. Unread #2 - Dec 3, 2009 at 8:55 AM
  4. A E G I Z
    Joined:
    Jun 7, 2009
    Posts:
    95
    Referrals:
    0
    Sythe Gold:
    0

    A E G I Z Member
    Banned

    paying algebra help equations!!

    Allright.

    1)3x^2+6x-8=0
    Discriminant = 4

    2) 5x^2+3x=12
    In standard: 5x^2+3x-12=0.
    Discriminant = Root from 69.

    3)7-5x^2+9x=x
    In standard: -5x^2+8x+7=0
    Discriminant = Root from 204.

    4)2x=x^2-x
    In standard: x^2+x=0
    Discriminant = 1

    5)x^2-3x+2=0
    Number of solutions = 2

    6)-3x^2+5x-1=0
    Number of solutions = 2

    7)x^2-2x+4=0
    Number of solutions = 1

    8)3x^2-6x+3=0
    Number of solutions = 1

    9)-5x^2+6x-6=0
    Number of solutions = 0

    I PM-ed you my PayPal.
     
  5. Unread #3 - Dec 24, 2009 at 9:57 AM
  6. A E G I Z
    Joined:
    Jun 7, 2009
    Posts:
    95
    Referrals:
    0
    Sythe Gold:
    0

    A E G I Z Member
    Banned

    paying algebra help equations!!

    Pay me or you're reported...
     
< Can't delet iPod touch apps | More Fake Vouch(es)/Ban Evaders >

Users viewing this thread
1 guest


 
 
Adblock breaks this site