H(x)=-16x^2+160x

Solution for H(x)=-16x^2+160x equation:

(H)=-16H^2+160H

We move all terms to the left:

(H)-(-16H^2+160H)=0

We get rid of parentheses

16H^2-160H+H=0

We add all the numbers together, and all the variables

16H^2-159H=0

a = 16; b = -159; c = 0;
Δ = b2-4ac
Δ = -1592-4·16·0
Δ = 25281
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25281}=159$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-159)-159}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-159)+159}{2*16}=\frac{318}{32}
=9+15/16
$