H(x)=-16x^2+32x+128

Solution for H(x)=-16x^2+32x+128 equation:

(H)=-16H^2+32H+128

We move all terms to the left:

(H)-(-16H^2+32H+128)=0

We get rid of parentheses

16H^2-32H+H-128=0

We add all the numbers together, and all the variables

16H^2-31H-128=0

a = 16; b = -31; c = -128;
Δ = b2-4ac
Δ = -312-4·16·(-128)
Δ = 9153
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{9153}=\sqrt{81*113}=\sqrt{81}*\sqrt{113}=9\sqrt{113}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-9\sqrt{113}}{2*16}=\frac{31-9\sqrt{113}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+9\sqrt{113}}{2*16}=\frac{31+9\sqrt{113}}{32}
$