(x+2)3x+(3x-2)(x+1)=2

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

(x+2)3x+(3x-2)(x+1)=2

We move all terms to the left:

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

We multiply parentheses

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2+7x-4=0

a = 6; b = 7; c = -4;
Δ = b2-4ac
Δ = 72-4·6·(-4)
Δ = 145
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{145}}{2*6}=\frac{-7-\sqrt{145}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{145}}{2*6}=\frac{-7+\sqrt{145}}{12}
$