0=12+64t-16t^2-32

Solution for 0=12+64t-16t^2-32 equation:

0=12+64t-16t^2-32

We move all terms to the left:

0-(12+64t-16t^2-32)=0

We add all the numbers together, and all the variables

-(12+64t-16t^2-32)=0

We get rid of parentheses

16t^2-64t-12+32=0

We add all the numbers together, and all the variables

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}
$