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

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

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

We move all terms to the left:

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

We get rid of parentheses

16H^2-80H+H-0=0

We add all the numbers together, and all the variables

16H^2-79H=0

a = 16; b = -79; c = 0;
Δ = b2-4ac
Δ = -792-4·16·0
Δ = 6241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6241}=79$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-79)-79}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-79)+79}{2*16}=\frac{158}{32}
=4+15/16
$