This is technically not a quadratic equation, but I am assuming you are trying to find the "zero's" of the function.
Anyhoozies, as you should know the generic formula for any quadratic problem is...
x = [-b ± √(b² - 4ac)] / 2a
In this case,
x = t
a = -16
b = -22
c = 220
So, we plug and chug these numbers in...
t = [-(-22) ± √((-22)² - 4(-16)(220))] / 2(-16)
When you solve this mess out (and it does get quite messy xD), you'll hopefully get
t = - (11 ± √3641) / 16
f(x)=-16t^2-22t+220
Solve using quadratic formula
1 answer