2=-16t^2+160t+2

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

2=-16t^2+160t+2

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

16t^2-160t=0

a = 16; b = -160; c = 0;
Δ = b2-4ac
Δ = -1602-4·16·0
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-160}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+160}{2*16}=\frac{320}{32}
=10
$