2007-12-19 15:34:52 · 6 answers · asked by AshconLivingston 2 in Science & Mathematics ➔ Mathematics
16 t ² - 160 t - 8 = 0
2 t ² - 20 t - 1 = 0
x = [ 20 ± √(400 + 8) ] / 4
x = [ 20 ± √(408) ] / 4
x = [ 20 ± 2√(102) ] / 4
x = 5 ± (1/2)√(102)
2007-12-19 21:13:00 · answer #1 · answered by Como 7 · 3⤊ 0⤋
The quadratic formula will work, but the numbers become more tractable if you divide everything by -8. This gives 2t^2 - 20t - 1 = 0, which doesn't have obvious factors, so we evaluate the quadratic formula to get ... you do it from here.
2007-12-19 15:41:13 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Okay sweet your doing Physics in your math class, this is the classic formula 1/2gt^2 + v(0)t + h(0) = h(f)
We are looking for when the thing hits the ground on it's way down :)
Easy way is to use the quadratic formula
-b +- sqroot of (b^2 - 4ac) all divide by 2a
t = -.04975 or t = 10.04975
2007-12-19 15:45:41 · answer #3 · answered by Timothy C 2 · 1⤊ 0⤋
[13]
-16t^2+160t+8=0
2t^2-20t-1=0[dividing both sides by -8]
t={20+-sqrt(400+8)}/4
=(20+-sqrt408)/4
=(20+-2sqrt102)/4
=5+-sqrt102/2
2007-12-19 15:45:37 · answer #4 · answered by alpha 7 · 1⤊ 0⤋
Eh I think you can simplify it to -8(2t^2 - 20t - 1)
2007-12-19 15:39:15 · answer #5 · answered by 港式奶茶 3 · 0⤊ 0⤋
used quadratic formula and got
t= -10 +-√51/-2
2007-12-19 15:44:44 · answer #6 · answered by Anonymous · 0⤊ 1⤋