1152=-16t^2+288t+0

Solution for 1152=-16t^2+288t+0 equation:

1152=-16t^2+288t+0

We move all terms to the left:

1152-(-16t^2+288t+0)=0

We get rid of parentheses

16t^2-288t-0+1152=0

We add all the numbers together, and all the variables

16t^2-288t+1152=0

a = 16; b = -288; c = +1152;
Δ = b2-4ac
Δ = -2882-4·16·1152
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9216}=96$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-288)-96}{2*16}=\frac{192}{32}
=6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-288)+96}{2*16}=\frac{384}{32}
=12
$