400(t)=-16t^2+160t+0

Solution for 400(t)=-16t^2+160t+0 equation:

400(t)=-16t^2+160t+0

We move all terms to the left:

400(t)-(-16t^2+160t+0)=0

We get rid of parentheses

16t^2-160t+400t-0=0

We add all the numbers together, and all the variables

16t^2+240t=0

a = 16; b = 240; c = 0;
Δ = b2-4ac
Δ = 2402-4·16·0
Δ = 57600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{57600}=240$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(240)-240}{2*16}=\frac{-480}{32}
=-15
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(240)+240}{2*16}=\frac{0}{32}
=0
$