H(t)=-16t^2+25t+0

Solution for H(t)=-16t^2+25t+0 equation:

(H)=-16H^2+25H+0

We move all terms to the left:

(H)-(-16H^2+25H+0)=0

We get rid of parentheses

16H^2-25H+H-0=0

We add all the numbers together, and all the variables

16H^2-24H=0

a = 16; b = -24; c = 0;
Δ = b2-4ac
Δ = -242-4·16·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-24}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+24}{2*16}=\frac{48}{32}
=1+1/2
$