-16t^2+72t+24=0

Solution for -16t^2+72t+24=0 equation:

-16t^2+72t+24=0

a = -16; b = 72; c = +24;
Δ = b2-4ac
Δ = 722-4·(-16)·24
Δ = 6720
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:

$\sqrt{\Delta}=\sqrt{6720}=\sqrt{64*105}=\sqrt{64}*\sqrt{105}=8\sqrt{105}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-8\sqrt{105}}{2*-16}=\frac{-72-8\sqrt{105}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+8\sqrt{105}}{2*-16}=\frac{-72+8\sqrt{105}}{-32}
$