x+(1/3x+100)+(1.25x-10)=400

Solution for x+(1/3x+100)+(1.25x-10)=400 equation:

x+(1/3x+100)+(1.25x-10)=400

We move all terms to the left:

x+(1/3x+100)+(1.25x-10)-(400)=0


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

x∈R


We get rid of parentheses

x+1/3x+1.25x+100-10-400=0

We multiply all the terms by the denominator


x*3x+(1.25x)*3x+100*3x-10*3x-400*3x+1=0

We add all the numbers together, and all the variables

x*3x+(+1.25x)*3x+100*3x-10*3x-400*3x+1=0

We multiply parentheses

3x^2+x*3x+100*3x-10*3x-400*3x+1=0

Wy multiply elements

3x^2+3x^2+300x-30x-1200x+1=0

We add all the numbers together, and all the variables

6x^2-930x+1=0

a = 6; b = -930; c = +1;
Δ = b2-4ac
Δ = -9302-4·6·1
Δ = 864876
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:

$\sqrt{\Delta}=\sqrt{864876}=\sqrt{4*216219}=\sqrt{4}*\sqrt{216219}=2\sqrt{216219}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-930)-2\sqrt{216219}}{2*6}=\frac{930-2\sqrt{216219}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-930)+2\sqrt{216219}}{2*6}=\frac{930+2\sqrt{216219}}{12}
$