3x-2/2x-3=3x-8/x+4

Solution for 3x-2/2x-3=3x-8/x+4 equation:

3x-2/2x-3=3x-8/x+4

We move all terms to the left:

3x-2/2x-3-(3x-8/x+4)=0


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R



Domain of the equation: x+4)!=0

x∈R


We get rid of parentheses

3x-2/2x-3x+8/x-4-3=0

We calculate fractions


3x-3x+(-2x)/2x^2+16x/2x^2-4-3=0

We add all the numbers together, and all the variables

(-2x)/2x^2+16x/2x^2-7=0

We multiply all the terms by the denominator


(-2x)+16x-7*2x^2=0

We add all the numbers together, and all the variables

16x+(-2x)-7*2x^2=0

Wy multiply elements

-14x^2+16x+(-2x)=0

We get rid of parentheses

-14x^2+16x-2x=0

We add all the numbers together, and all the variables

-14x^2+14x=0

a = -14; b = 14; c = 0;
Δ = b2-4ac
Δ = 142-4·(-14)·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14}{2*-14}=\frac{-28}{-28}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14}{2*-14}=\frac{0}{-28}
=0
$