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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-32t+12=0

a = 16; b = -32; c = +12;
Δ = b2-4ac
Δ = -322-4·16·12
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16}{2*16}=\frac{16}{32}
=1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16}{2*16}=\frac{48}{32}
=1+1/2
$