-24t^2+80t-24=0

Solution for -24t^2+80t-24=0 equation:

-24t^2+80t-24=0

a = -24; b = 80; c = -24;
Δ = b2-4ac
Δ = 802-4·(-24)·(-24)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-64}{2*-24}=\frac{-144}{-48}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+64}{2*-24}=\frac{-16}{-48}
=1/3
$