5v^2+29v+20=0
6n^2-19n+15=0
2k^2+7k=0
3b^2+b-10=0
2007-04-04 02:08:31 · 4 answers · asked by The Dark Side of The Force 3 in Science & Mathematics ➔ Mathematics
5v^2+29v+20=0
(5v+4)(v+5) = 0
v = -5 and -4/5
6n^2-19n+15=0
(2n - 3)(3n - 5) = 0
n = 3/2 and 5/3
2k^2+7k=0
k(2k + 7) = 0
k = 0 and -7/2
3b^2+b-10=0
(3b - 5)(b + 2) = 0
b = -2 and 5/3
2007-04-04 02:21:12 · answer #1 · answered by Mathematica 7 · 2⤊ 0⤋
5v^2+29v+20=0
v= -0.8 or v = -5
6n^2-19n+15=0
n=1.66 or n=1.5
2k^2+7k=0
k= 0 or k = -3.5
3b^2+b-10=0
b= 1.66 or b= -2
2007-04-04 09:30:44 · answer #2 · answered by edison c d 4 · 0⤊ 0⤋
5v² + 29v + 20 = 0
5v² + 25v + 4v + 20 = 0
5x(v + 5) + 4(v + 5) = 0
(5v + 4)(v + 5) = 0
- - - - - - -
6n² - 19n + 15= 0
6n² - 9n - 10n + 15 = 0
3n(2n - 3) -5(2n - 3)
(3n - 5)(2n - 3)
- - - - - - - - -s-
2007-04-04 10:10:22 · answer #3 · answered by SAMUEL D 7 · 0⤊ 0⤋
(1) v=-5, v=-0.8
(2) n=(5/3), n=1.5
(3) k=0, k=-3.5
(4) b=-2, b=(5/3)
Please give me best answer thanks!
2007-04-04 09:47:22 · answer #4 · answered by Anonymous · 0⤊ 0⤋