h(t) = 13 + 16t - 16t^2?

Question

solve for t.

i reeeally don't understand this factoring stuff -_-

Answer 1

Follow my step below:

h(t) = 13 + 16t - 16t²

Let say, h(t) = 0
So, 13 + 16t - 16t² = 0

Next, find a, b, and c:
-16t² + 16t + 13 = 0
So, a = -16, b = 16, and c = 13

Solve it by using the formula:

t = (-b ± √ b² - 4ac) ٪ 2a
t = (-16 ± √ (16)² - 4(-16)(13)) ٪ 2(-16)
t = (-16 ± √ (256 + 832)) ٪ -32
t = (-16 ± √ (1088)) ٪ -32
t = (-16 ± 32.984845) ٪ -32

First answer:
t = (-16 + 32.984845) ٪ -32
t = -0.53078

Second answer:
t = (-16 - 32.984845) ٪ -32
t = 1.53078

Therefore, the answer must be:
t = -0.53078 or t = 1.53078

I hope you get the best answer for your question. Happy Halloween's Day too..

Answer 2

just use quadratic formula

since h(t) is the same as y, you will have

y = 13 +16t - 16t^2
y = -16t^2 + 16t + 13

quadratic formula is at^2 + bt + c = 0, the value of x is given by:

t = [ -b ± sqrt(b^2 - 4ac) ] / 2a

just place the numbers and your x result will be: t = -[(sqrt17)-2] / 4 or t= [(sqrt 17)+2] / 4

Answer 3

To factor, you would have to rearrange things so
0 = (13-H(t))+16t - 16t^2. 13-H(t) is a constant.
Then you could use the quadratic equation.

Answer 4

i dont think ms. exclusive is right. if u have (8t - 13)(2t +1) you would get 16t^2 - 18t-13

Answer 5

(8t - 13)(2t +1)=0

8t - 13 = 0 and 2t + 1 =0
8t = 13 and 2t = -1
t = 13/8 and t = -1/2
t = 1.625 and t = -0.5