2=400-16t^2

Solution for 2=400-16t^2 equation:

2=400-16t^2

We move all terms to the left:

2-(400-16t^2)=0

We get rid of parentheses

16t^2-400+2=0

We add all the numbers together, and all the variables

16t^2-398=0

a = 16; b = 0; c = -398;
Δ = b2-4ac
Δ = 02-4·16·(-398)
Δ = 25472
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:

$\sqrt{\Delta}=\sqrt{25472}=\sqrt{64*398}=\sqrt{64}*\sqrt{398}=8\sqrt{398}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{398}}{2*16}=\frac{0-8\sqrt{398}}{32}
=-\frac{8\sqrt{398}}{32}
=-\frac{\sqrt{398}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{398}}{2*16}=\frac{0+8\sqrt{398}}{32}
=\frac{8\sqrt{398}}{32}
=\frac{\sqrt{398}}{4}
$