3/4x-6+2x=7/4x-3

Solution for 3/4x-6+2x=7/4x-3 equation:

3/4x-6+2x=7/4x-3

We move all terms to the left:

3/4x-6+2x-(7/4x-3)=0


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R



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

x∈R


We add all the numbers together, and all the variables

2x+3/4x-(7/4x-3)-6=0

We get rid of parentheses

2x+3/4x-7/4x+3-6=0

We multiply all the terms by the denominator


2x*4x+3*4x-6*4x+3-7=0

We add all the numbers together, and all the variables

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

Wy multiply elements

8x^2+12x-24x-4=0

We add all the numbers together, and all the variables

8x^2-12x-4=0

a = 8; b = -12; c = -4;
Δ = b2-4ac
Δ = -122-4·8·(-4)
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{17}}{2*8}=\frac{12-4\sqrt{17}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{17}}{2*8}=\frac{12+4\sqrt{17}}{16}
$