8/3x+(x-20)+x=180

Solution for 8/3x+(x-20)+x=180 equation:

8/3x+(x-20)+x=180

We move all terms to the left:

8/3x+(x-20)+x-(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

x+8/3x+(x-20)-180=0

We get rid of parentheses

x+8/3x+x-20-180=0

We multiply all the terms by the denominator


x*3x+x*3x-20*3x-180*3x+8=0

Wy multiply elements

3x^2+3x^2-60x-540x+8=0

We add all the numbers together, and all the variables

6x^2-600x+8=0

a = 6; b = -600; c = +8;
Δ = b2-4ac
Δ = -6002-4·6·8
Δ = 359808
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:

$\sqrt{\Delta}=\sqrt{359808}=\sqrt{64*5622}=\sqrt{64}*\sqrt{5622}=8\sqrt{5622}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-600)-8\sqrt{5622}}{2*6}=\frac{600-8\sqrt{5622}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-600)+8\sqrt{5622}}{2*6}=\frac{600+8\sqrt{5622}}{12}
$