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November 25, 2014

November 25, 2014

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

(the 2 is an exponent)

idk how 2 do it

the teacher told us how but didnt show us

plz help

- math -
**Reiny**, Wednesday, March 23, 2011 at 10:20pm6n^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 -
**Angel**, Wednesday, March 23, 2011 at 10:20pmty

- math -
**Jai**, Wednesday, March 23, 2011 at 10:29pm6n^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 -
**Bella**, Wednesday, March 23, 2011 at 10:34pmlol 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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