6/x+1-(3/2)=8/3x+3

Solution for 6/x+1-(3/2)=8/3x+3 equation:

6/x+1-(3/2)=8/3x+3

We move all terms to the left:

6/x+1-(3/2)-(8/3x+3)=0


Domain of the equation: x!=0

x∈R



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

x∈R


We add all the numbers together, and all the variables

6/x-(8/3x+3)+1-(+3/2)=0

We get rid of parentheses

6/x-8/3x-3+1-3/2=0

We calculate fractions


(-27x^2)/12x^2+72x/12x^2+(-32x)/12x^2-3+1=0

We add all the numbers together, and all the variables

(-27x^2)/12x^2+72x/12x^2+(-32x)/12x^2-2=0

We multiply all the terms by the denominator


(-27x^2)+72x+(-32x)-2*12x^2=0

Wy multiply elements

(-27x^2)-24x^2+72x+(-32x)=0

We get rid of parentheses

-27x^2-24x^2+72x-32x=0

We add all the numbers together, and all the variables

-51x^2+40x=0

a = -51; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·(-51)·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*-51}=\frac{-80}{-102}
=40/51
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*-51}=\frac{0}{-102}
=0
$