(2x-100)+(x-20)+(1/4x+40)=180

Solution for (2x-100)+(x-20)+(1/4x+40)=180 equation:

(2x-100)+(x-20)+(1/4x+40)=180

We move all terms to the left:

(2x-100)+(x-20)+(1/4x+40)-(180)=0


Domain of the equation: 4x+40)!=0

x∈R


We get rid of parentheses

2x+x+1/4x-100-20+40-180=0

We multiply all the terms by the denominator


2x*4x+x*4x-100*4x-20*4x+40*4x-180*4x+1=0

Wy multiply elements

8x^2+4x^2-400x-80x+160x-720x+1=0

We add all the numbers together, and all the variables

12x^2-1040x+1=0

a = 12; b = -1040; c = +1;
Δ = b2-4ac
Δ = -10402-4·12·1
Δ = 1081552
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:

$\sqrt{\Delta}=\sqrt{1081552}=\sqrt{16*67597}=\sqrt{16}*\sqrt{67597}=4\sqrt{67597}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1040)-4\sqrt{67597}}{2*12}=\frac{1040-4\sqrt{67597}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1040)+4\sqrt{67597}}{2*12}=\frac{1040+4\sqrt{67597}}{24}
$