-16t^2+96t-44=0

Solution for -16t^2+96t-44=0 equation:

-16t^2+96t-44=0

a = -16; b = 96; c = -44;
Δ = b2-4ac
Δ = 962-4·(-16)·(-44)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-80}{2*-16}=\frac{-176}{-32}
=5+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+80}{2*-16}=\frac{-16}{-32}
=1/2
$