-16t^2+6t+24=0

Solution for -16t^2+6t+24=0 equation:

-16t^2+6t+24=0

a = -16; b = 6; c = +24;
Δ = b2-4ac
Δ = 62-4·(-16)·24
Δ = 1572
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:

$\sqrt{\Delta}=\sqrt{1572}=\sqrt{4*393}=\sqrt{4}*\sqrt{393}=2\sqrt{393}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{393}}{2*-16}=\frac{-6-2\sqrt{393}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{393}}{2*-16}=\frac{-6+2\sqrt{393}}{-32}
$