1/3x+(x-10)+(x-20)=320

Solution for 1/3x+(x-10)+(x-20)=320 equation:

1/3x+(x-10)+(x-20)=320

We move all terms to the left:

1/3x+(x-10)+(x-20)-(320)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

3x^2+3x^2-30x-60x-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}
$