6+3x(2x+1)=3x+12

Solution for 6+3x(2x+1)=3x+12 equation:

6+3x(2x+1)=3x+12

We move all terms to the left:

6+3x(2x+1)-(3x+12)=0

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2-6=0

a = 6; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·6·(-6)
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{-(0)-12}{2*6}=\frac{-12}{12}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*6}=\frac{12}{12}
=1
$