-16t2+288t=0

Solution for -16t2+288t=0 equation:

-16t^2+288t=0

a = -16; b = 288; c = 0;
Δ = b2-4ac
Δ = 2882-4·(-16)·0
Δ = 82944
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{82944}=288$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(288)-288}{2*-16}=\frac{-576}{-32}
=+18
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(288)+288}{2*-16}=\frac{0}{-32}
=0
$