-16t2+40t-16=0

Solution for -16t2+40t-16=0 equation:

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

a = -16; b = 40; c = -16;
Δ = b2-4ac
Δ = 402-4·(-16)·(-16)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-24}{2*-16}=\frac{-64}{-32}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+24}{2*-16}=\frac{-16}{-32}
=1/2
$