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
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
To solve the quadratic equation 6n^2 - 19n + 15 = 0, you can use the quadratic formula. The quadratic formula is:
n = (-b ± √(b^2 - 4ac)) / (2a)
In your equation, a = 6, b = -19, and c = 15. Plugging these values into the quadratic formula, we get:
n = (-(-19) ± √((-19)^2 - 4 * 6 * 15)) / (2 * 6)
Simplifying further:
n = (19 ± √(361 - 360)) / 12
n = (19 ± √1) / 12
To find the solutions, we calculate both possibilities:
n1 = (19 + 1) / 12 = 20 / 12 = 5/3
n2 = (19 - 1) / 12 = 18 / 12 = 3/2
Therefore, the solutions to the equation 6n^2 - 19n + 15 = 0 are n = 5/3 and n = 3/2.
You can check these solutions by plugging them back into the original equation and verifying that both sides of the equation are equal.