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

Solution for H(t)=-16t^2+4t equation:

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

We move all terms to the left:

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

We get rid of parentheses

16H^2-4H+H=0

We add all the numbers together, and all the variables

16H^2-3H=0

a = 16; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·16·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*16}=\frac{6}{32}
=3/16
$