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

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

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

We move all terms to the left:

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

We multiply parentheses

6x-(3x(x+4))+8=0


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

3x(x+4)

We multiply parentheses

3x^2+12x

Back to the equation:

-(3x^2+12x)


We get rid of parentheses

-3x^2+6x-12x+8=0

We add all the numbers together, and all the variables

-3x^2-6x+8=0

a = -3; b = -6; c = +8;
Δ = b2-4ac
Δ = -62-4·(-3)·8
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{33}}{2*-3}=\frac{6-2\sqrt{33}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{33}}{2*-3}=\frac{6+2\sqrt{33}}{-6}
$