(4x-100)+(x+20)+(1/5x+2)=180

Solution for (4x-100)+(x+20)+(1/5x+2)=180 equation:

(4x-100)+(x+20)+(1/5x+2)=180

We move all terms to the left:

(4x-100)+(x+20)+(1/5x+2)-(180)=0


Domain of the equation: 5x+2)!=0

x∈R


We get rid of parentheses

4x+x+1/5x-100+20+2-180=0

We multiply all the terms by the denominator


4x*5x+x*5x-100*5x+20*5x+2*5x-180*5x+1=0

Wy multiply elements

20x^2+5x^2-500x+100x+10x-900x+1=0

We add all the numbers together, and all the variables

25x^2-1290x+1=0

a = 25; b = -1290; c = +1;
Δ = b2-4ac
Δ = -12902-4·25·1
Δ = 1664000
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:

$\sqrt{\Delta}=\sqrt{1664000}=\sqrt{25600*65}=\sqrt{25600}*\sqrt{65}=160\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1290)-160\sqrt{65}}{2*25}=\frac{1290-160\sqrt{65}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1290)+160\sqrt{65}}{2*25}=\frac{1290+160\sqrt{65}}{50}
$