-16t^2+64t-20=0

Solution for -16t^2+64t-20=0 equation:

-16t^2+64t-20=0

a = -16; b = 64; c = -20;
Δ = b2-4ac
Δ = 642-4·(-16)·(-20)
Δ = 2816
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:

$\sqrt{\Delta}=\sqrt{2816}=\sqrt{256*11}=\sqrt{256}*\sqrt{11}=16\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-16\sqrt{11}}{2*-16}=\frac{-64-16\sqrt{11}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+16\sqrt{11}}{2*-16}=\frac{-64+16\sqrt{11}}{-32}
$