-16t^2+96t+144=0

Solution for -16t^2+96t+144=0 equation:

-16t^2+96t+144=0

a = -16; b = 96; c = +144;
Δ = b2-4ac
Δ = 962-4·(-16)·144
Δ = 18432
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:

$\sqrt{\Delta}=\sqrt{18432}=\sqrt{9216*2}=\sqrt{9216}*\sqrt{2}=96\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-96\sqrt{2}}{2*-16}=\frac{-96-96\sqrt{2}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+96\sqrt{2}}{2*-16}=\frac{-96+96\sqrt{2}}{-32}
$