0=4+64t-16t^2-32

Solution for 0=4+64t-16t^2-32 equation:

0=4+64t-16t^2-32

We move all terms to the left:

0-(4+64t-16t^2-32)=0

We add all the numbers together, and all the variables

-(4+64t-16t^2-32)=0

We get rid of parentheses

16t^2-64t-4+32=0

We add all the numbers together, and all the variables

16t^2-64t+28=0

a = 16; b = -64; c = +28;
Δ = b2-4ac
Δ = -642-4·16·28
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-48}{2*16}=\frac{16}{32}
=1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+48}{2*16}=\frac{112}{32}
=3+1/2
$