H(x)=4x+4/2x+5

Solution for H(x)=4x+4/2x+5 equation:

(H)=4H+4/2H+5

We move all terms to the left:

(H)-(4H+4/2H+5)=0


Domain of the equation: 2H+5)!=0

H∈R


We get rid of parentheses

H-4H-4/2H-5=0

We multiply all the terms by the denominator


H*2H-4H*2H-5*2H-4=0

Wy multiply elements

2H^2-8H^2-10H-4=0

We add all the numbers together, and all the variables

-6H^2-10H-4=0

a = -6; b = -10; c = -4;
Δ = b2-4ac
Δ = -102-4·(-6)·(-4)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2}{2*-6}=\frac{8}{-12}
=-2/3
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2}{2*-6}=\frac{12}{-12}
=-1
$