(2/3x)+(x+40)=180

Solution for (2/3x)+(x+40)=180 equation:

(2/3x)+(x+40)=180

We move all terms to the left:

(2/3x)+(x+40)-(180)=0


Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+2/3x)+(x+40)-180=0

We get rid of parentheses

2/3x+x+40-180=0

We multiply all the terms by the denominator


x*3x+40*3x-180*3x+2=0

Wy multiply elements

3x^2+120x-540x+2=0

We add all the numbers together, and all the variables

3x^2-420x+2=0

a = 3; b = -420; c = +2;
Δ = b2-4ac
Δ = -4202-4·3·2
Δ = 176376
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:

$\sqrt{\Delta}=\sqrt{176376}=\sqrt{4*44094}=\sqrt{4}*\sqrt{44094}=2\sqrt{44094}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-420)-2\sqrt{44094}}{2*3}=\frac{420-2\sqrt{44094}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-420)+2\sqrt{44094}}{2*3}=\frac{420+2\sqrt{44094}}{6}
$