3x+2-1/3(4x)=2+2x

Solution for 3x+2-1/3(4x)=2+2x equation:

3x+2-1/3(4x)=2+2x

We move all terms to the left:

3x+2-1/3(4x)-(2+2x)=0


Domain of the equation: 34x!=0

x!=0/34

x!=0

x∈R


We add all the numbers together, and all the variables

3x-1/34x-(2x+2)+2=0

We get rid of parentheses

3x-1/34x-2x-2+2=0

We multiply all the terms by the denominator


3x*34x-2x*34x-2*34x+2*34x-1=0

Wy multiply elements

102x^2-68x^2-68x+68x-1=0

We add all the numbers together, and all the variables

34x^2-1=0

a = 34; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·34·(-1)
Δ = 136
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:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{34}}{2*34}=\frac{0-2\sqrt{34}}{68}
=-\frac{2\sqrt{34}}{68}
=-\frac{\sqrt{34}}{34}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{34}}{2*34}=\frac{0+2\sqrt{34}}{68}
=\frac{2\sqrt{34}}{68}
=\frac{\sqrt{34}}{34}
$