x+(1/3x+100)+(1.25x-10)=120

Solution for x+(1/3x+100)+(1.25x-10)=120 equation:

x+(1/3x+100)+(1.25x-10)=120

We move all terms to the left:

x+(1/3x+100)+(1.25x-10)-(120)=0


Domain of the equation: 3x+100)!=0

x∈R


We get rid of parentheses

x+1/3x+1.25x+100-10-120=0

We multiply all the terms by the denominator


x*3x+(1.25x)*3x+100*3x-10*3x-120*3x+1=0

We add all the numbers together, and all the variables

x*3x+(+1.25x)*3x+100*3x-10*3x-120*3x+1=0

We multiply parentheses

3x^2+x*3x+100*3x-10*3x-120*3x+1=0

Wy multiply elements

3x^2+3x^2+300x-30x-360x+1=0

We add all the numbers together, and all the variables

6x^2-90x+1=0

a = 6; b = -90; c = +1;
Δ = b2-4ac
Δ = -902-4·6·1
Δ = 8076
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:

$\sqrt{\Delta}=\sqrt{8076}=\sqrt{4*2019}=\sqrt{4}*\sqrt{2019}=2\sqrt{2019}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-90)-2\sqrt{2019}}{2*6}=\frac{90-2\sqrt{2019}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-90)+2\sqrt{2019}}{2*6}=\frac{90+2\sqrt{2019}}{12}
$