Posted by Angel on Wednesday, March 23, 2011 at 10:13pm.
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
ty
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~ :)
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