0=-16t^2+80t+96
Solve for t.
Can someone please show me the steps on how to do this I can not seem to get it right.
Thanks!!
imagine at^2+bt+c
t = ( -b +- sqrt((b^2)-(4ac)) )/ 2a
you will have to possible answer
0= -16( t^2 -5t -6)
0 = -16(t+ 1)(t-6)
t + 1 = 0
t + 1 -1 = 0-1
t = -1 Does not make sense you can ignore when t is -1
t -6 = 0
t -6 + 6 = 6
t = 6
t = 6 seconds
To solve the equation 0 = -16t^2 + 80t + 96, you can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using:
x = (-b ± √(b^2 - 4ac)) / (2a)
In the given equation, a = -16, b = 80, and c = 96. Let's substitute these values into the quadratic formula:
t = (-80 ± √((80)^2 - 4(-16)(96))) / (2(-16))
Simplifying further:
t = (-80 ± √(6400 + 6144)) / (-32)
t = (-80 ± √(12544)) / (-32)
t = (-80 ± √(112^2)) / (-32)
The square root of 112 can be simplified as 8√7:
t = (-80 ± 8√7) / (-32)
Now, let's simplify further:
t = (80 ± 8√7) / 32
t = 10 ± √7/4
So, the solutions for t are:
t = 10 + √7/4
t = 10 - √7/4
Therefore, the solutions for t are approximately 1.18 and 18.82.
To solve the quadratic equation 0 = -16t^2 + 80t + 96, you can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In your equation, a = -16, b = 80, and c = 96. Substituting these values into the formula, we get:
t = (-80 ± √(80^2 - 4(-16)(96))) / (2(-16))
Simplifying further:
t = (-80 ± √(6400 + 6144)) / (-32)
t = (-80 ± √(12544)) / (-32)
t = (-80 ± 112) / (-32)
Now, we have two possible solutions:
1. t = (-80 + 112) / (-32) = 32/(-32) = -1
2. t = (-80 - 112) / (-32) = -192/(-32) = 6
Therefore, the solutions of the equation are t = -1 and t = 6.