0=2+160t-16t^2

Solution for 0=2+160t-16t^2 equation:

0=2+160t-16t^2

We move all terms to the left:

0-(2+160t-16t^2)=0

We add all the numbers together, and all the variables

-(2+160t-16t^2)=0

We get rid of parentheses

16t^2-160t-2=0

a = 16; b = -160; c = -2;
Δ = b2-4ac
Δ = -1602-4·16·(-2)
Δ = 25728
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:

$\sqrt{\Delta}=\sqrt{25728}=\sqrt{64*402}=\sqrt{64}*\sqrt{402}=8\sqrt{402}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-8\sqrt{402}}{2*16}=\frac{160-8\sqrt{402}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+8\sqrt{402}}{2*16}=\frac{160+8\sqrt{402}}{32}
$