0=288+96t-16t^2

Solution for 0=288+96t-16t^2 equation:

0=288+96t-16t^2

We move all terms to the left:

0-(288+96t-16t^2)=0

We add all the numbers together, and all the variables

-(288+96t-16t^2)=0

We get rid of parentheses

16t^2-96t-288=0

a = 16; b = -96; c = -288;
Δ = b2-4ac
Δ = -962-4·16·(-288)
Δ = 27648
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{27648}=\sqrt{9216*3}=\sqrt{9216}*\sqrt{3}=96\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-96\sqrt{3}}{2*16}=\frac{96-96\sqrt{3}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+96\sqrt{3}}{2*16}=\frac{96+96\sqrt{3}}{32}
$