0=-16t2+48t-32

Solution for 0=-16t2+48t-32 equation:

0=-16t^2+48t-32

We move all terms to the left:

0-(-16t^2+48t-32)=0

We add all the numbers together, and all the variables

-(-16t^2+48t-32)=0

We get rid of parentheses

16t^2-48t+32=0

a = 16; b = -48; c = +32;
Δ = b2-4ac
Δ = -482-4·16·32
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{-(-48)-16}{2*16}=\frac{32}{32}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16}{2*16}=\frac{64}{32}
=2
$