3x(4x+2)+3(x+4)=12x+6x+3x+12

Solution for 3x(4x+2)+3(x+4)=12x+6x+3x+12 equation:

3x(4x+2)+3(x+4)=12x+6x+3x+12

We move all terms to the left:

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

We add all the numbers together, and all the variables

3x(4x+2)+3(x+4)-(21x+12)=0

We multiply parentheses

12x^2+6x+3x-(21x+12)+12=0

We get rid of parentheses

12x^2+6x+3x-21x-12+12=0

We add all the numbers together, and all the variables

12x^2-12x=0

a = 12; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·12·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*12}=\frac{0}{24}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*12}=\frac{24}{24}
=1
$