2006-08-11 04:52:03 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
it is n not y
2006-08-11 05:00:39 · update #1
Did you ask for y ?
Well, in this particular question, y can take any value.
But if it's asked for n, then n=1.5 or 2.
2006-08-11 04:59:04 · answer #1 · answered by Samvit 1 · 1⤊ 1⤋
2n² - 7n +6 = 0 factors into
(2n-3)*(n-2) = 0 so
n = 3/2 or n=2
Doug
2006-08-11 05:02:23 · answer #2 · answered by doug_donaghue 7 · 0⤊ 0⤋
2n^2 - 7n + 6 = 0
(2n - 3)(n - 2) = 0
2n - 3 = 0
2n = 3
n = 3/2; or n = 2
2006-08-11 17:42:53 · answer #3 · answered by ronw 4 · 0⤊ 0⤋
2n²-7n+6=0
(2n-3)(n-2)=0
Thus:
2n-3=0 or n-2=0
2n=1 or n=2
n=3*½ or n=2
2006-08-11 05:01:30 · answer #4 · answered by Victor C 3 · 0⤊ 0⤋
3/2 or 1/4
2006-08-11 07:26:44 · answer #5 · answered by sundar 2 · 0⤊ 0⤋
2n^2-4n-3n+6=0
(2n-3) (n-2)
n=3/2 or 2
2006-08-11 05:00:06 · answer #6 · answered by raj 7 · 0⤊ 0⤋
2n^2 - 7n + 6 = 0
n = (-b ± sqrt(b^2 - 4ac))/(2a)
n = (-(-7) ± sqrt((-7)^2 - 4(2)(6)))/(2(2))
n = (7 ± sqrt(49 - 48))/4
n = (7 ± sqrt(1))/4
n = (7 ± 1)/4
n = (8/4) or (6/4)
n = 2 or (3/2)
ANS : n = 2 or (3/2)
2006-08-11 13:03:59 · answer #7 · answered by Sherman81 6 · 0⤊ 0⤋
4 - 4y - y^2 = 0 -y^2 - 4y + 4 = 0 -(y^2 + 4y - 4) = 0 y^2 - 4y - 4 = 0 y = (-b ± sqrt(b^2 - 4ac))/(2a) y = (-(-4) ± sqrt((-4)^2 - 4(a million)(-4)))/(2(a million)) y = (4 ± sqrt(16 + 16))/2 y = (4 ± sqrt(32))/2 y = (4 ± sqrt(16 * 2))/2 y = (4 ± 4sqrt(2))/2 ANS : 2 + 2sqrt(2) or 2 - 2sqrt(2)
2016-11-24 20:10:25 · answer #8 · answered by parvin 4 · 0⤊ 0⤋