2/3x+8=2x+4

Solution for 2/3x+8=2x+4 equation:

2/3x+8=2x+4

We move all terms to the left:

2/3x+8-(2x+4)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We get rid of parentheses

2/3x-2x-4+8=0

We multiply all the terms by the denominator


-2x*3x-4*3x+8*3x+2=0

Wy multiply elements

-6x^2-12x+24x+2=0

We add all the numbers together, and all the variables

-6x^2+12x+2=0

a = -6; b = 12; c = +2;
Δ = b2-4ac
Δ = 122-4·(-6)·2
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{3}}{2*-6}=\frac{-12-8\sqrt{3}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{3}}{2*-6}=\frac{-12+8\sqrt{3}}{-12}
$