factor
2006-11-05 07:21:39 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
t^2+t+12 = 0
t=[- 1 +- sqrt(1-4(12 ) ]/2
which means that there is no real solution,
so the polynomial cannot be factored at all
s
2006-11-05 07:44:31 · answer #1 · answered by Anonymous · 0⤊ 1⤋
You cant factor this, you might be able to if it was t^2+t-12 which would be (t+4)(t-3)
2006-11-05 15:27:47 · answer #2 · answered by kiyomidog 2 · 1⤊ 0⤋
t^2+t+12=0 ==> Delta=1^2 - 4(1)(12)= -47
so,there are no answers for this equation.
2006-11-05 15:40:20 · answer #3 · answered by Alex 4 · 1⤊ 0⤋
x1 = (-b + square_root((b^2) - 4ac)) / (2a)
= (-1 + square_root(1^2 - 4.1.12)) / (2.1)
= (-1 + square_root(-47)) / 2
= (-1 + square_root(47)(i)) / 2
= -(1/2) + (i/2).square_root(47)
x2 = (-b - square_root((b^2) - 4ac)) / (2a)
= (-1 - square_root(1^2 - 4.1.12)) / (2.1)
= (-1 - square_root(-47)) / 2
= (-1 - square_root(47)(i)) / 2
= -(1/2) - (i/2).square_root(47)
i = square_root(-1)
2006-11-05 15:54:57 · answer #4 · answered by d@mn 2 · 1⤊ 0⤋
-47i/2
47i/2
2006-11-05 15:46:20 · answer #5 · answered by oee22 2 · 1⤊ 0⤋