3=-16t2+160t+2

Solution for 3=-16t2+160t+2 equation:

3=-16t^2+160t+2

We move all terms to the left:

3-(-16t^2+160t+2)=0

We get rid of parentheses

16t^2-160t-2+3=0

We add all the numbers together, and all the variables

16t^2-160t+1=0

a = 16; b = -160; c = +1;
Δ = b2-4ac
Δ = -1602-4·16·1
Δ = 25536
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:

$\sqrt{\Delta}=\sqrt{25536}=\sqrt{64*399}=\sqrt{64}*\sqrt{399}=8\sqrt{399}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-8\sqrt{399}}{2*16}=\frac{160-8\sqrt{399}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+8\sqrt{399}}{2*16}=\frac{160+8\sqrt{399}}{32}
$