2x+6/4x+3=3x-2/4x+5

Solution for 2x+6/4x+3=3x-2/4x+5 equation:

2x+6/4x+3=3x-2/4x+5

We move all terms to the left:

2x+6/4x+3-(3x-2/4x+5)=0


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R



Domain of the equation: 4x+5)!=0

x∈R


We get rid of parentheses

2x+6/4x-3x+2/4x-5+3=0

We multiply all the terms by the denominator


2x*4x-3x*4x-5*4x+3*4x+6+2=0

We add all the numbers together, and all the variables

2x*4x-3x*4x-5*4x+3*4x+8=0

Wy multiply elements

8x^2-12x^2-20x+12x+8=0

We add all the numbers together, and all the variables

-4x^2-8x+8=0

a = -4; b = -8; c = +8;
Δ = b2-4ac
Δ = -82-4·(-4)·8
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{3}}{2*-4}=\frac{8-8\sqrt{3}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{3}}{2*-4}=\frac{8+8\sqrt{3}}{-8}
$