-16t^2+64t-32=0

Solution for -16t^2+64t-32=0 equation:

-16t^2+64t-32=0

a = -16; b = 64; c = -32;
Δ = b2-4ac
Δ = 642-4·(-16)·(-32)
Δ = 2048
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:

$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-32\sqrt{2}}{2*-16}=\frac{-64-32\sqrt{2}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+32\sqrt{2}}{2*-16}=\frac{-64+32\sqrt{2}}{-32}
$