-16t^2+24t+5=0

Solution for -16t^2+24t+5=0 equation:

-16t^2+24t+5=0

a = -16; b = 24; c = +5;
Δ = b2-4ac
Δ = 242-4·(-16)·5
Δ = 896
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:

$\sqrt{\Delta}=\sqrt{896}=\sqrt{64*14}=\sqrt{64}*\sqrt{14}=8\sqrt{14}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{14}}{2*-16}=\frac{-24-8\sqrt{14}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{14}}{2*-16}=\frac{-24+8\sqrt{14}}{-32}
$