2x+(1/3x-10)=4

Solution for 2x+(1/3x-10)=4 equation:

2x+(1/3x-10)=4

We move all terms to the left:

2x+(1/3x-10)-(4)=0


Domain of the equation: 3x-10)!=0

x∈R


We get rid of parentheses

2x+1/3x-10-4=0

We multiply all the terms by the denominator


2x*3x-10*3x-4*3x+1=0

Wy multiply elements

6x^2-30x-12x+1=0

We add all the numbers together, and all the variables

6x^2-42x+1=0

a = 6; b = -42; c = +1;
Δ = b2-4ac
Δ = -422-4·6·1
Δ = 1740
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:

$\sqrt{\Delta}=\sqrt{1740}=\sqrt{4*435}=\sqrt{4}*\sqrt{435}=2\sqrt{435}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{435}}{2*6}=\frac{42-2\sqrt{435}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{435}}{2*6}=\frac{42+2\sqrt{435}}{12}
$