3/4x+6=2/3x-4

Solution for 3/4x+6=2/3x-4 equation:

3/4x+6=2/3x-4

We move all terms to the left:

3/4x+6-(2/3x-4)=0


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R



Domain of the equation: 3x-4)!=0

x∈R


We get rid of parentheses

3/4x-2/3x+4+6=0

We calculate fractions


9x/12x^2+(-8x)/12x^2+4+6=0

We add all the numbers together, and all the variables

9x/12x^2+(-8x)/12x^2+10=0

We multiply all the terms by the denominator


9x+(-8x)+10*12x^2=0

Wy multiply elements

120x^2+9x+(-8x)=0

We get rid of parentheses

120x^2+9x-8x=0

We add all the numbers together, and all the variables

120x^2+x=0

a = 120; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·120·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*120}=\frac{-2}{240}
=-1/120
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*120}=\frac{0}{240}
=0
$