-16t^2+192t-512=0

Solution for -16t^2+192t-512=0 equation:

-16t^2+192t-512=0

a = -16; b = 192; c = -512;
Δ = b2-4ac
Δ = 1922-4·(-16)·(-512)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(192)-64}{2*-16}=\frac{-256}{-32}
=+8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(192)+64}{2*-16}=\frac{-128}{-32}
=+4
$