H(t)^-1=-16t^2+128t

Solution for H(t)^-1=-16t^2+128t equation:

(H)^-1=-16H^2+128H

We move all terms to the left:

(H)^-1-(-16H^2+128H)=0

We add all the numbers together, and all the variables

-(-16H^2+128H)+H-1=0

We get rid of parentheses

16H^2-128H+H-1=0

We add all the numbers together, and all the variables

16H^2-127H-1=0

a = 16; b = -127; c = -1;
Δ = b2-4ac
Δ = -1272-4·16·(-1)
Δ = 16193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-127)-\sqrt{16193}}{2*16}=\frac{127-\sqrt{16193}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-127)+\sqrt{16193}}{2*16}=\frac{127+\sqrt{16193}}{32}
$