-96t+80t-16t^2=0

Solution for -96t+80t-16t^2=0 equation:

-96t+80t-16t^2=0

We add all the numbers together, and all the variables

-16t^2-16t=0

a = -16; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·(-16)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*-16}=\frac{0}{-32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*-16}=\frac{32}{-32}
=-1
$