(X-20)+40+(x-10)+1/3x=360

Solution for (X-20)+40+(x-10)+1/3x=360 equation:

(X-20)+40+(X-10)+1/3X=360

We move all terms to the left:

(X-20)+40+(X-10)+1/3X-(360)=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-320=0

We get rid of parentheses

X+X+1/3X-20-10-320=0

We multiply all the terms by the denominator


X*3X+X*3X-20*3X-10*3X-320*3X+1=0

Wy multiply elements

3X^2+3X^2-60X-30X-960X+1=0

We add all the numbers together, and all the variables

6X^2-1050X+1=0

a = 6; b = -1050; c = +1;
Δ = b2-4ac
Δ = -10502-4·6·1
Δ = 1102476
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:

$\sqrt{\Delta}=\sqrt{1102476}=\sqrt{4*275619}=\sqrt{4}*\sqrt{275619}=2\sqrt{275619}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1050)-2\sqrt{275619}}{2*6}=\frac{1050-2\sqrt{275619}}{12}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1050)+2\sqrt{275619}}{2*6}=\frac{1050+2\sqrt{275619}}{12}
$