40+(x-20)+(x-10)-1/3x=36

Solution for 40+(x-20)+(x-10)-1/3x=36 equation:

40+(x-20)+(x-10)-1/3x=36

We move all terms to the left:

40+(x-20)+(x-10)-1/3x-(36)=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-20)+(x-10)-1/3x+4=0

We get rid of parentheses

x+x-1/3x-20-10+4=0

We multiply all the terms by the denominator


x*3x+x*3x-20*3x-10*3x+4*3x-1=0

Wy multiply elements

3x^2+3x^2-60x-30x+12x-1=0

We add all the numbers together, and all the variables

6x^2-78x-1=0

a = 6; b = -78; c = -1;
Δ = b2-4ac
Δ = -782-4·6·(-1)
Δ = 6108
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:

$\sqrt{\Delta}=\sqrt{6108}=\sqrt{4*1527}=\sqrt{4}*\sqrt{1527}=2\sqrt{1527}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-78)-2\sqrt{1527}}{2*6}=\frac{78-2\sqrt{1527}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-78)+2\sqrt{1527}}{2*6}=\frac{78+2\sqrt{1527}}{12}
$