-16t^2+143t+24=0

Solution for -16t^2+143t+24=0 equation:

-16t^2+143t+24=0

a = -16; b = 143; c = +24;
Δ = b2-4ac
Δ = 1432-4·(-16)·24
Δ = 21985
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(143)-\sqrt{21985}}{2*-16}=\frac{-143-\sqrt{21985}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(143)+\sqrt{21985}}{2*-16}=\frac{-143+\sqrt{21985}}{-32}
$