H=-16t2+4t+0

Solution for H=-16t2+4t+0 equation:

=-16H^2+4H+0

We move all terms to the left:

-(-16H^2+4H+0)=0

We get rid of parentheses

16H^2-4H-0=0

We add all the numbers together, and all the variables

16H^2-4H=0

a = 16; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·16·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*16}=\frac{8}{32}
=1/4
$