2x+32/3x+40=12/3x+10

Solution for 2x+32/3x+40=12/3x+10 equation:

2x+32/3x+40=12/3x+10

We move all terms to the left:

2x+32/3x+40-(12/3x+10)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



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

x∈R


We get rid of parentheses

2x+32/3x-12/3x-10+40=0

We multiply all the terms by the denominator


2x*3x-10*3x+40*3x+32-12=0

We add all the numbers together, and all the variables

2x*3x-10*3x+40*3x+20=0

Wy multiply elements

6x^2-30x+120x+20=0

We add all the numbers together, and all the variables

6x^2+90x+20=0

a = 6; b = 90; c = +20;
Δ = b2-4ac
Δ = 902-4·6·20
Δ = 7620
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:

$\sqrt{\Delta}=\sqrt{7620}=\sqrt{4*1905}=\sqrt{4}*\sqrt{1905}=2\sqrt{1905}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-2\sqrt{1905}}{2*6}=\frac{-90-2\sqrt{1905}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+2\sqrt{1905}}{2*6}=\frac{-90+2\sqrt{1905}}{12}
$