H(x)=-16x^2+120x

Solution for H(x)=-16x^2+120x equation:

(H)=-16H^2+120H

We move all terms to the left:

(H)-(-16H^2+120H)=0

We get rid of parentheses

16H^2-120H+H=0

We add all the numbers together, and all the variables

16H^2-119H=0

a = 16; b = -119; c = 0;
Δ = b2-4ac
Δ = -1192-4·16·0
Δ = 14161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14161}=119$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-119)-119}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-119)+119}{2*16}=\frac{238}{32}
=7+7/16
$