1/3x+(x+5)+x=40

Solution for 1/3x+(x+5)+x=40 equation:

1/3x+(x+5)+x=40

We move all terms to the left:

1/3x+(x+5)+x-(40)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

x+1/3x+(x+5)-40=0

We get rid of parentheses

x+1/3x+x+5-40=0

We multiply all the terms by the denominator


x*3x+x*3x+5*3x-40*3x+1=0

Wy multiply elements

3x^2+3x^2+15x-120x+1=0

We add all the numbers together, and all the variables

6x^2-105x+1=0

a = 6; b = -105; c = +1;
Δ = b2-4ac
Δ = -1052-4·6·1
Δ = 11001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-105)-\sqrt{11001}}{2*6}=\frac{105-\sqrt{11001}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-105)+\sqrt{11001}}{2*6}=\frac{105+\sqrt{11001}}{12}
$