(x-10)+1/3x+(x-20)+4=360

Solution for (x-10)+1/3x+(x-20)+4=360 equation:

(x-10)+1/3x+(x-20)+4=360

We move all terms to the left:

(x-10)+1/3x+(x-20)+4-(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-10)+1/3x+(x-20)-356=0

We get rid of parentheses

x+1/3x+x-10-20-356=0

We multiply all the terms by the denominator


x*3x+x*3x-10*3x-20*3x-356*3x+1=0

Wy multiply elements

3x^2+3x^2-30x-60x-1068x+1=0

We add all the numbers together, and all the variables

6x^2-1158x+1=0

a = 6; b = -1158; c = +1;
Δ = b2-4ac
Δ = -11582-4·6·1
Δ = 1340940
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:

$\sqrt{\Delta}=\sqrt{1340940}=\sqrt{4*335235}=\sqrt{4}*\sqrt{335235}=2\sqrt{335235}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1158)-2\sqrt{335235}}{2*6}=\frac{1158-2\sqrt{335235}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1158)+2\sqrt{335235}}{2*6}=\frac{1158+2\sqrt{335235}}{12}
$