-384+160t-16t^2=0

Solution for -384+160t-16t^2=0 equation:

-384+160t-16t^2=0

a = -16; b = 160; c = -384;
Δ = b2-4ac
Δ = 1602-4·(-16)·(-384)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-32}{2*-16}=\frac{-192}{-32}
=+6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+32}{2*-16}=\frac{-128}{-32}
=+4
$