Question

MEDAL!!!

Answer 1

\[\huge 3n^2 - 11 = 8n \]

Answer 2

solve for n?

Answer 3

yes using quadratic formula

Answer 4

\[\large 3n^2 - 8n - 11 = 0\]
would that equation be correct?

Answer 5

yes

Answer 6

use quadratic formula

\[\frac{ -b \pm \sqrt{b^2-4ac }}{ 2a }\]

where ax^2+bx+c=0

yes this equation is correct

substitute values:

Answer 7

\[n = \frac{ 8 \pm \sqrt{64 - 4(-33)} }{ 6 }\]

Answer 8

looks good

Answer 9

yeah correct

now simplify this

Answer 10

\[n = \frac{ 8 \pm \sqrt{196} }{ 8 } = \frac{ 8 + 14 }{ 8 }\]

Answer 11

how denominator became 8 from 6?

Answer 12

how did your 6 become an 8 in the denominator

Answer 13

everything is correct exept the denominator

Answer 14

sorry typo :\

Answer 15

ok

Answer 16

ah ;)

Answer 17

it would be \[\frac{ 8 \pm 14 }{ ? }\]

Answer 18

woops 6 instead of ?

Answer 19

(3.66, -1)

Answer 20

correct

Answer 21

thank you @Lily2913 @AlexandervonHumboldt2

Answer 22

np :)

Answer 23

np :)