320t-16t^2=0

Solution for 320t-16t^2=0 equation:

320t-16t^2=0

a = -16; b = 320; c = 0;
Δ = b2-4ac
Δ = 3202-4·(-16)·0
Δ = 102400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{102400}=320$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(320)-320}{2*-16}=\frac{-640}{-32}
=+20
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(320)+320}{2*-16}=\frac{0}{-32}
=0
$