4/3x+x=40

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

4/3x+x=40

We move all terms to the left:

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


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

x+4/3x-40=0

We multiply all the terms by the denominator


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

Wy multiply elements

3x^2-120x+4=0

a = 3; b = -120; c = +4;
Δ = b2-4ac
Δ = -1202-4·3·4
Δ = 14352
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:

$\sqrt{\Delta}=\sqrt{14352}=\sqrt{16*897}=\sqrt{16}*\sqrt{897}=4\sqrt{897}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-4\sqrt{897}}{2*3}=\frac{120-4\sqrt{897}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+4\sqrt{897}}{2*3}=\frac{120+4\sqrt{897}}{6}
$