2200t-16t^2+320t=0

Solution for 2200t-16t^2+320t=0 equation:

2200t-16t^2+320t=0

We add all the numbers together, and all the variables

-16t^2+2520t=0

a = -16; b = 2520; c = 0;
Δ = b2-4ac
Δ = 25202-4·(-16)·0
Δ = 6350400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6350400}=2520$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2520)-2520}{2*-16}=\frac{-5040}{-32}
=157+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2520)+2520}{2*-16}=\frac{0}{-32}
=0
$