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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3x^2+2x-10=0

a = 3; b = 2; c = -10;
Δ = b2-4ac
Δ = 22-4·3·(-10)
Δ = 124
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:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{31}}{2*3}=\frac{-2-2\sqrt{31}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{31}}{2*3}=\frac{-2+2\sqrt{31}}{6}
$