my answer is -3,7
2007-11-28 06:50:24 · 7 answers · asked by thonia 1 in Science & Mathematics ➔ Mathematics
t^2 - 4t - 21 = 0
(t-7)(t+3)=0
t = 7 and -3
so you are correct.
2007-11-28 06:55:22 · answer #1 · answered by norman 7 · 1⤊ 0⤋
Solve 21=t^2-4t?
t^2 - 4t - 21 = 0
(t - 7) * (t + 3) = 0
t - 7 = 0; t + 3 = 0
t = 7; t = -3
2007-11-28 07:20:22 · answer #2 · answered by Anonymous · 1⤊ 0⤋
OK
21=t^2-4t
0 = t^2 - 4t - 21
0 = (t -7)(t+3)
t = 7 and -3
I agree with you!!
Hope that helps.
2007-11-28 06:56:47 · answer #3 · answered by pyz01 7 · 1⤊ 0⤋
One way to do it is to make it into the quadratic format:
0 = t^2 - 4t - 21
And now factor.
0 = (t - 7)(t+3)
Therefore, t = 7, -3.
Another way would be to use the quadratic formula, but this one had a simple factor.
2007-11-28 06:57:04 · answer #4 · answered by Mary 2 · 1⤊ 0⤋
t^2-4t-21=0
(t+3)(t-7)=0
t=-3 or 7
Your answer is correct!
2007-11-28 10:04:23 · answer #5 · answered by Anonymous · 0⤊ 0⤋
21=t^2-4t
0=t^2-4t-21
0=t^2-7t+3t-21
0=t(t-7)+3(t-7)
0=(t-7)(t+3)
so t-7=0
t=7
or
t+3=0
t=-3
so t =7, -3
ur answer is right
2007-11-28 07:03:52 · answer #6 · answered by Siva 5 · 1⤊ 0⤋
t^2-7t+3t-21=0
t(t-7)+3(t-7)=0
(t-7) (t+3)
t=7, t=-3
2007-11-28 07:34:05 · answer #7 · answered by surmy 2 · 1⤊ 0⤋