-16t^2+62t+24=0

Solution for -16t^2+62t+24=0 equation:

-16t^2+62t+24=0

a = -16; b = 62; c = +24;
Δ = b2-4ac
Δ = 622-4·(-16)·24
Δ = 5380
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:

$\sqrt{\Delta}=\sqrt{5380}=\sqrt{4*1345}=\sqrt{4}*\sqrt{1345}=2\sqrt{1345}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(62)-2\sqrt{1345}}{2*-16}=\frac{-62-2\sqrt{1345}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62)+2\sqrt{1345}}{2*-16}=\frac{-62+2\sqrt{1345}}{-32}
$