96=16t^2+80t

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

96=16t^2+80t

We move all terms to the left:

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

We get rid of parentheses

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

a = -16; b = -80; c = +96;
Δ = b2-4ac
Δ = -802-4·(-16)·96
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-112}{2*-16}=\frac{-32}{-32}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+112}{2*-16}=\frac{192}{-32}
=-6
$