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Posted by on Wednesday, March 23, 2011 at 10:13pm.

6n2-19n+15=o

(the 2 is an exponent)
idk how 2 do it
the teacher told us how but didn't show us
plz help

  • math - , Wednesday, March 23, 2011 at 10:20pm

    6n^2 - 19n + 15 = 0
    factors to
    (3x-5)(2x-3) = 0
    so 3x-5 = 0 or 2x-3 = 0
    x = 5/3 or x = 3/2

  • math - , Wednesday, March 23, 2011 at 10:20pm

    ty

  • math - , Wednesday, March 23, 2011 at 10:29pm

    6n^2-19n+15=0
    to solve this we can either factor it (if it's factorable) or use quadratic formula,, here we just use quadratic formula:
    for a given quadratic equation, ax^2 + bx + c = 0,
    x = [-b +- sqrt(b^2 - 4*a*c)]/(2*a)
    note: +- is plus or minus
    where
    a = numerical coeff of x^2,
    b = numerical coeff of x and
    c = constant
    in the problem,
    6n^2-19n+15=0 (note that x is just replaced by the x variable)
    a = 6
    b = -19
    c = 15
    then we substitute it to quadratic formula:
    x = [-b +- sqrt(b^2 - 4*a*c)]/(2*a)
    x = [-(-19) +- sqrt((-19)^2 - 4*6*15)]/(2*6)
    we can separate it into plus and minus:
    using plus:
    x = [-(-19) + sqrt((-19)^2 - 4*6*15)]/(2*6)
    x = [19 + sqrt(361-360)]/(12)
    x = [19 + sqrt(1)]/12
    x = (19 + 1)/12
    x = 5/3 (this is the first root)

    for the minus:
    x = [-(-19) - sqrt((-19)^2 - 4*6*15)]/(2*6)
    x = [19 - sqrt(1)]/12
    x = 18/12
    x = 3/2 (second root)

    *note: variable x here is also n

    hope this helps~ :)

  • math - , Wednesday, March 23, 2011 at 10:34pm

    lol thx both of yall an srry but i change names cuz ppl alwayz think angel is my real name an its not so i change it 2 make sure ppl dnt do that

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