H(x)=2/3x+4

Solution for H(x)=2/3x+4 equation:

(H)=2/3H+4

We move all terms to the left:

(H)-(2/3H+4)=0


Domain of the equation: 3H+4)!=0

H∈R


We get rid of parentheses

H-2/3H-4=0

We multiply all the terms by the denominator


H*3H-4*3H-2=0

Wy multiply elements

3H^2-12H-2=0

a = 3; b = -12; c = -2;
Δ = b2-4ac
Δ = -122-4·3·(-2)
Δ = 168
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{42}}{2*3}=\frac{12-2\sqrt{42}}{6}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{42}}{2*3}=\frac{12+2\sqrt{42}}{6}
$