0=48+32t-16t2

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

0=48+32t-16t^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-32t-48=0

a = 16; b = -32; c = -48;
Δ = b2-4ac
Δ = -322-4·16·(-48)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-64}{2*16}=\frac{-32}{32}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+64}{2*16}=\frac{96}{32}
=3
$