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

Solution for 80(t)=-16t^2+0 equation:

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

We move all terms to the left:

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

We get rid of parentheses

16t^2+80t-0=0

We add all the numbers together, and all the variables

16t^2+80t=0

a = 16; b = 80; c = 0;
Δ = b2-4ac
Δ = 802-4·16·0
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-80}{2*16}=\frac{-160}{32}
=-5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+80}{2*16}=\frac{0}{32}
=0
$