H(x)=(4/5)(x^2+12x-64)

Solution for H(x)=(4/5)(x^2+12x-64) equation:

(H)=(4/5)(H^2+12H-64)

We move all terms to the left:

(H)-((4/5)(H^2+12H-64))=0


Domain of the equation: 5)(H^2+12H-64))!=0

H∈R


We add all the numbers together, and all the variables

H-((+4/5)(H^2+12H-64))=0

We multiply all the terms by the denominator


H*5)(H^2+12H+4-64))-((=0

We add all the numbers together, and all the variables

12H+H*5)(H^2=0

Wy multiply elements

5H^2+12H=0

a = 5; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·5·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*5}=\frac{-24}{10}
=-2+2/5
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*5}=\frac{0}{10}
=0
$