(x+4)(3x-2)-(3x-1)(3x+1)=2x(11-3x)

Solution for (x+4)(3x-2)-(3x-1)(3x+1)=2x(11-3x) equation:

(x+4)(3x-2)-(3x-1)(3x+1)=2x(11-3x)

We move all terms to the left:

(x+4)(3x-2)-(3x-1)(3x+1)-(2x(11-3x))=0

We add all the numbers together, and all the variables

(x+4)(3x-2)-(3x-1)(3x+1)-(2x(-3x+11))=0

We use the square of the difference formula

9x^2+(x+4)(3x-2)-(2x(-3x+11))+1=0

We multiply parentheses ..

9x^2+(+3x^2-2x+12x-8)-(2x(-3x+11))+1=0


We calculate terms in parentheses: -(2x(-3x+11)), so:

2x(-3x+11)

We multiply parentheses

-6x^2+22x

Back to the equation:

-(-6x^2+22x)


We get rid of parentheses

9x^2+3x^2+6x^2-2x+12x-22x-8+1=0

We add all the numbers together, and all the variables

18x^2-12x-7=0

a = 18; b = -12; c = -7;
Δ = b2-4ac
Δ = -122-4·18·(-7)
Δ = 648
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:

$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-18\sqrt{2}}{2*18}=\frac{12-18\sqrt{2}}{36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+18\sqrt{2}}{2*18}=\frac{12+18\sqrt{2}}{36}
$