(1/3x)+(x)+(x+20)=132

Solution for (1/3x)+(x)+(x+20)=132 equation:

(1/3x)+(x)+(x+20)=132

We move all terms to the left:

(1/3x)+(x)+(x+20)-(132)=0


Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+1/3x)+x+(x+20)-132=0

We add all the numbers together, and all the variables

x+(+1/3x)+(x+20)-132=0

We get rid of parentheses

x+1/3x+x+20-132=0

We multiply all the terms by the denominator


x*3x+x*3x+20*3x-132*3x+1=0

Wy multiply elements

3x^2+3x^2+60x-396x+1=0

We add all the numbers together, and all the variables

6x^2-336x+1=0

a = 6; b = -336; c = +1;
Δ = b2-4ac
Δ = -3362-4·6·1
Δ = 112872
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:

$\sqrt{\Delta}=\sqrt{112872}=\sqrt{4*28218}=\sqrt{4}*\sqrt{28218}=2\sqrt{28218}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-336)-2\sqrt{28218}}{2*6}=\frac{336-2\sqrt{28218}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-336)+2\sqrt{28218}}{2*6}=\frac{336+2\sqrt{28218}}{12}
$