0=-16t^2+224t

Solution for 0=-16t^2+224t equation:

0=-16t^2+224t

We move all terms to the left:

0-(-16t^2+224t)=0

We add all the numbers together, and all the variables

-(-16t^2+224t)=0

We get rid of parentheses

16t^2-224t=0

a = 16; b = -224; c = 0;
Δ = b2-4ac
Δ = -2242-4·16·0
Δ = 50176
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{50176}=224$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-224)-224}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-224)+224}{2*16}=\frac{448}{32}
=14
$