1/3x+(x-20)+(x-10)+40=180

Solution for 1/3x+(x-20)+(x-10)+40=180 equation:

1/3x+(x-20)+(x-10)+40=180

We move all terms to the left:

1/3x+(x-20)+(x-10)+40-(180)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

1/3x+(x-20)+(x-10)-140=0

We get rid of parentheses

1/3x+x+x-20-10-140=0

We multiply all the terms by the denominator


x*3x+x*3x-20*3x-10*3x-140*3x+1=0

Wy multiply elements

3x^2+3x^2-60x-30x-420x+1=0

We add all the numbers together, and all the variables

6x^2-510x+1=0

a = 6; b = -510; c = +1;
Δ = b2-4ac
Δ = -5102-4·6·1
Δ = 260076
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:

$\sqrt{\Delta}=\sqrt{260076}=\sqrt{4*65019}=\sqrt{4}*\sqrt{65019}=2\sqrt{65019}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-510)-2\sqrt{65019}}{2*6}=\frac{510-2\sqrt{65019}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-510)+2\sqrt{65019}}{2*6}=\frac{510+2\sqrt{65019}}{12}
$