6*(3x+12)=2x*(3x+12)

Solution for 6*(3x+12)=2x*(3x+12) equation:

6(3x+12)=2x(3x+12)

We move all terms to the left:

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

We multiply parentheses

18x-(2x(3x+12))+72=0


We calculate terms in parentheses: -(2x(3x+12)), so:

2x(3x+12)

We multiply parentheses

6x^2+24x

Back to the equation:

-(6x^2+24x)


We get rid of parentheses

-6x^2+18x-24x+72=0

We add all the numbers together, and all the variables

-6x^2-6x+72=0

a = -6; b = -6; c = +72;
Δ = b2-4ac
Δ = -62-4·(-6)·72
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-42}{2*-6}=\frac{-36}{-12}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+42}{2*-6}=\frac{48}{-12}
=-4
$