(2/3c)-2=2+c

Solution for (2/3c)-2=2+c equation:

(2/3c)-2=2+c

We move all terms to the left:

(2/3c)-2-(2+c)=0


Domain of the equation: 3c)!=0

c!=0/1

c!=0

c∈R


We add all the numbers together, and all the variables

(+2/3c)-(c+2)-2=0

We get rid of parentheses

2/3c-c-2-2=0

We multiply all the terms by the denominator


-c*3c-2*3c-2*3c+2=0

Wy multiply elements

-3c^2-6c-6c+2=0

We add all the numbers together, and all the variables

-3c^2-12c+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:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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