0=320-16t^2-0

Solution for 0=320-16t^2-0 equation:

0=320-16t^2-0

We move all terms to the left:

0-(320-16t^2-0)=0

We add all the numbers together, and all the variables

-(320-16t^2-0)=0

We get rid of parentheses

16t^2-320+0=0

We add all the numbers together, and all the variables

16t^2-320=0

a = 16; b = 0; c = -320;
Δ = b2-4ac
Δ = 02-4·16·(-320)
Δ = 20480
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:

$\sqrt{\Delta}=\sqrt{20480}=\sqrt{4096*5}=\sqrt{4096}*\sqrt{5}=64\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64\sqrt{5}}{2*16}=\frac{0-64\sqrt{5}}{32}
=-\frac{64\sqrt{5}}{32}
=-2\sqrt{5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64\sqrt{5}}{2*16}=\frac{0+64\sqrt{5}}{32}
=\frac{64\sqrt{5}}{32}
=2\sqrt{5}
$