H(x)=16x^2+80x

Solution for H(x)=16x^2+80x equation:

(H)=16H^2+80H

We move all terms to the left:

(H)-(16H^2+80H)=0

We get rid of parentheses

-16H^2+H-80H=0

We add all the numbers together, and all the variables

-16H^2-79H=0

a = -16; b = -79; c = 0;
Δ = b2-4ac
Δ = -792-4·(-16)·0
Δ = 6241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6241}=79$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-79)-79}{2*-16}=\frac{0}{-32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-79)+79}{2*-16}=\frac{158}{-32}
=-4+15/16
$