0=-64t^2+32t+4

Solution for 0=-64t^2+32t+4 equation:

0=-64t^2+32t+4

We move all terms to the left:

0-(-64t^2+32t+4)=0

We add all the numbers together, and all the variables

-(-64t^2+32t+4)=0

We get rid of parentheses

64t^2-32t-4=0

a = 64; b = -32; c = -4;
Δ = b2-4ac
Δ = -322-4·64·(-4)
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{-(-32)-32\sqrt{2}}{2*64}=\frac{32-32\sqrt{2}}{128}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+32\sqrt{2}}{2*64}=\frac{32+32\sqrt{2}}{128}
$