H(t)=16t2+128t

Solution for H(t)=16t2+128t equation:

(H)=16H^2+128H

We move all terms to the left:

(H)-(16H^2+128H)=0

We get rid of parentheses

-16H^2+H-128H=0

We add all the numbers together, and all the variables

-16H^2-127H=0

a = -16; b = -127; c = 0;
Δ = b2-4ac
Δ = -1272-4·(-16)·0
Δ = 16129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16129}=127$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-127)-127}{2*-16}=\frac{0}{-32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-127)+127}{2*-16}=\frac{254}{-32}
=-7+15/16
$