34x^2 + 406x +1180 = 0
2007-07-15 01:56:45 · 5 answers · asked by sweet_candy 2 in Science & Mathematics ➔ Mathematics
34x^2 + 406x +1180 = 0
2(17x^2 + 203x + 590) = 0
2(17x^2 + 85x + 118x + 590) = 0
2(17x(x+5) + 118(x+5)) = 0
2(x+5)(17x+118) = 0
x = -5
x = -118/17 = -6.94 (approx.)
x = {-118/17, -5}
2007-07-15 02:04:08 · answer #1 · answered by gudspeling 7 · 1⤊ 1⤋
34x^2 + 406x + 1180 = 0
Divide both sides by 2,
17x^2 + 203x + 590 = 0
17x^2 + 85x + 118x + 590 = 0
17x(x + 5) + 118(x + 5) = 0
(17x + 118)(x + 5) = 0
17x + 118 = 0 (or) x + 5 = 0
17x = -118 (or) x = -5
x = -118/17 (or) x = -5
2007-07-15 09:23:05 · answer #2 · answered by Akilesh - Internet Undertaker 7 · 2⤊ 0⤋
divide each number by 2. Then factor out that answer. You get (17x+59) (x+10)
2007-07-15 12:45:45 · answer #3 · answered by Kandice F 4 · 0⤊ 0⤋
2(17x+118)(x+5)
2007-07-15 09:05:45 · answer #4 · answered by Wayne M 2 · 1⤊ 0⤋
(x +6.941176471 )( x+ 5)
2007-07-15 09:02:22 · answer #5 · answered by yang 4 · 0⤊ 0⤋