0=280-16t^2

Solution for 0=280-16t^2 equation:

0=280-16t^2

We move all terms to the left:

0-(280-16t^2)=0

We add all the numbers together, and all the variables

-(280-16t^2)=0

We get rid of parentheses

16t^2-280=0

a = 16; b = 0; c = -280;
Δ = b2-4ac
Δ = 02-4·16·(-280)
Δ = 17920
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:

$\sqrt{\Delta}=\sqrt{17920}=\sqrt{256*70}=\sqrt{256}*\sqrt{70}=16\sqrt{70}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{70}}{2*16}=\frac{0-16\sqrt{70}}{32}
=-\frac{16\sqrt{70}}{32}
=-\frac{\sqrt{70}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{70}}{2*16}=\frac{0+16\sqrt{70}}{32}
=\frac{16\sqrt{70}}{32}
=\frac{\sqrt{70}}{2}
$