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

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

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

We add all the numbers together, and all the variables

-16t^2+80t=0

a = -16; b = 80; c = 0;
Δ = b2-4ac
Δ = 802-4·(-16)·0
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{-(80)-80}{2*-16}=\frac{-160}{-32}
=+5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+80}{2*-16}=\frac{0}{-32}
=0
$