2x+4/3x=40

Solution for 2x+4/3x=40 equation:

2x+4/3x=40

We move all terms to the left:

2x+4/3x-(40)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We multiply all the terms by the denominator


2x*3x-40*3x+4=0

Wy multiply elements

6x^2-120x+4=0

a = 6; b = -120; c = +4;
Δ = b2-4ac
Δ = -1202-4·6·4
Δ = 14304
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:

$\sqrt{\Delta}=\sqrt{14304}=\sqrt{16*894}=\sqrt{16}*\sqrt{894}=4\sqrt{894}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-4\sqrt{894}}{2*6}=\frac{120-4\sqrt{894}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+4\sqrt{894}}{2*6}=\frac{120+4\sqrt{894}}{12}
$