24=16.t2+40.t

Solution for 24=16.t2+40.t equation:

24=16.t^2+40.t

We move all terms to the left:

24-(16.t^2+40.t)=0

We get rid of parentheses

-16.t^2-40.t+24=0

We add all the numbers together, and all the variables

-16t^2-40t+24=0

a = -16; b = -40; c = +24;
Δ = b2-4ac
Δ = -402-4·(-16)·24
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-56}{2*-16}=\frac{-16}{-32}
=1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+56}{2*-16}=\frac{96}{-32}
=-3
$