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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

2(x-3)x

We multiply parentheses

2x^2-6x

Back to the equation:

-(2x^2-6x)


We get rid of parentheses

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

We add all the numbers together, and all the variables

4x^2+18x=0

a = 4; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·4·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*4}=\frac{-36}{8}
=-4+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*4}=\frac{0}{8}
=0
$