200=2+160t-16t^2

Solution for 200=2+160t-16t^2 equation:

200=2+160t-16t^2

We move all terms to the left:

200-(2+160t-16t^2)=0

We get rid of parentheses

16t^2-160t-2+200=0

We add all the numbers together, and all the variables

16t^2-160t+198=0

a = 16; b = -160; c = +198;
Δ = b2-4ac
Δ = -1602-4·16·198
Δ = 12928
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:

$\sqrt{\Delta}=\sqrt{12928}=\sqrt{64*202}=\sqrt{64}*\sqrt{202}=8\sqrt{202}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-8\sqrt{202}}{2*16}=\frac{160-8\sqrt{202}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+8\sqrt{202}}{2*16}=\frac{160+8\sqrt{202}}{32}
$