0=(-16t^2)+560t

Solution for 0=(-16t^2)+560t equation:

0=(-16t^2)+560t

We move all terms to the left:

0-((-16t^2)+560t)=0

We add all the numbers together, and all the variables

-((-16t^2)+560t)=0


We calculate terms in parentheses: -((-16t^2)+560t), so:

(-16t^2)+560t

We get rid of parentheses

-16t^2+560t

Back to the equation:

-(-16t^2+560t)


We get rid of parentheses

16t^2-560t=0

a = 16; b = -560; c = 0;
Δ = b2-4ac
Δ = -5602-4·16·0
Δ = 313600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{313600}=560$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-560)-560}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-560)+560}{2*16}=\frac{1120}{32}
=35
$