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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


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

4x(3x+1)

We multiply parentheses

12x^2+4x

Back to the equation:

-(12x^2+4x)


We get rid of parentheses

-12x^2+12x-4x+6=0

We add all the numbers together, and all the variables

-12x^2+8x+6=0

a = -12; b = 8; c = +6;
Δ = b2-4ac
Δ = 82-4·(-12)·6
Δ = 352
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:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{22}}{2*-12}=\frac{-8-4\sqrt{22}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{22}}{2*-12}=\frac{-8+4\sqrt{22}}{-24}
$