2(1/3r)+1/3r+r=324

Solution for 2(1/3r)+1/3r+r=324 equation:

2(1/3r)+1/3r+r=324

We move all terms to the left:

2(1/3r)+1/3r+r-(324)=0


Domain of the equation: 3r)!=0

r!=0/1

r!=0

r∈R



Domain of the equation: 3r!=0

r!=0/3

r!=0

r∈R


We add all the numbers together, and all the variables

2(+1/3r)+1/3r+r-324=0

We add all the numbers together, and all the variables

r+2(+1/3r)+1/3r-324=0

We multiply parentheses

r+2r+1/3r-324=0

We multiply all the terms by the denominator


r*3r+2r*3r-324*3r+1=0

Wy multiply elements

3r^2+6r^2-972r+1=0

We add all the numbers together, and all the variables

9r^2-972r+1=0

a = 9; b = -972; c = +1;
Δ = b2-4ac
Δ = -9722-4·9·1
Δ = 944748
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{944748}=\sqrt{36*26243}=\sqrt{36}*\sqrt{26243}=6\sqrt{26243}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-972)-6\sqrt{26243}}{2*9}=\frac{972-6\sqrt{26243}}{18}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-972)+6\sqrt{26243}}{2*9}=\frac{972+6\sqrt{26243}}{18}
$