3x(6x+4)=(2x-1)(3x+1)

Solution for 3x(6x+4)=(2x-1)(3x+1) equation:

3x(6x+4)=(2x-1)(3x+1)

We move all terms to the left:

3x(6x+4)-((2x-1)(3x+1))=0

We multiply parentheses

18x^2+12x-((2x-1)(3x+1))=0

We multiply parentheses ..

18x^2-((+6x^2+2x-3x-1))+12x=0


We calculate terms in parentheses: -((+6x^2+2x-3x-1)), so:

(+6x^2+2x-3x-1)

We get rid of parentheses

6x^2+2x-3x-1

We add all the numbers together, and all the variables

6x^2-1x-1

Back to the equation:

-(6x^2-1x-1)


We add all the numbers together, and all the variables

18x^2+12x-(6x^2-1x-1)=0

We get rid of parentheses

18x^2-6x^2+12x+1x+1=0

We add all the numbers together, and all the variables

12x^2+13x+1=0

a = 12; b = 13; c = +1;
Δ = b2-4ac
Δ = 132-4·12·1
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*12}=\frac{-24}{24}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*12}=\frac{-2}{24}
=-1/12
$