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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

2x+12x-(3x(4x+3)+1)+8-6=0


We calculate terms in parentheses: -(3x(4x+3)+1), so:

3x(4x+3)+1

We multiply parentheses

12x^2+9x+1

Back to the equation:

-(12x^2+9x+1)


We add all the numbers together, and all the variables

14x-(12x^2+9x+1)+2=0

We get rid of parentheses

-12x^2+14x-9x-1+2=0

We add all the numbers together, and all the variables

-12x^2+5x+1=0

a = -12; b = 5; c = +1;
Δ = b2-4ac
Δ = 52-4·(-12)·1
Δ = 73
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{-(5)-\sqrt{73}}{2*-12}=\frac{-5-\sqrt{73}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{73}}{2*-12}=\frac{-5+\sqrt{73}}{-24}
$