-0.8t^2+16t=0

Solution for -0.8t^2+16t=0 equation:

-0.8t^2+16t=0

a = -0.8; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-0.8)·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*-0.8}=\frac{-32}{-1.6}
=+20
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-0.8}=\frac{0}{-1.6}
=0
$