0=-16t^2+125t+24

Solution for 0=-16t^2+125t+24 equation:

0=-16t^2+125t+24

We move all terms to the left:

0-(-16t^2+125t+24)=0

We add all the numbers together, and all the variables

-(-16t^2+125t+24)=0

We get rid of parentheses

16t^2-125t-24=0

a = 16; b = -125; c = -24;
Δ = b2-4ac
Δ = -1252-4·16·(-24)
Δ = 17161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{17161}=131$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-125)-131}{2*16}=\frac{-6}{32}
=-3/16
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-125)+131}{2*16}=\frac{256}{32}
=8
$