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

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

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

We move all terms to the left:

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


Domain of the equation: 7x!=0

x!=0/7

x!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


2x+4x+(-36x-54)/42x+(-49x)/42x+3=0

We add all the numbers together, and all the variables

6x+(-36x-54)/42x+(-49x)/42x+3=0

We multiply all the terms by the denominator


6x*42x+(-36x-54)+(-49x)+3*42x=0

Wy multiply elements

252x^2+(-36x-54)+(-49x)+126x=0

We get rid of parentheses

252x^2-36x-49x+126x-54=0

We add all the numbers together, and all the variables

252x^2+41x-54=0

a = 252; b = 41; c = -54;
Δ = b2-4ac
Δ = 412-4·252·(-54)
Δ = 56113
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-\sqrt{56113}}{2*252}=\frac{-41-\sqrt{56113}}{504}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+\sqrt{56113}}{2*252}=\frac{-41+\sqrt{56113}}{504}
$