-16t^2+40t-1=0

Solution for -16t^2+40t-1=0 equation:

-16t^2+40t-1=0

a = -16; b = 40; c = -1;
Δ = b2-4ac
Δ = 402-4·(-16)·(-1)
Δ = 1536
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-16\sqrt{6}}{2*-16}=\frac{-40-16\sqrt{6}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+16\sqrt{6}}{2*-16}=\frac{-40+16\sqrt{6}}{-32}
$