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

Solution for (3x+6)(x+3)+(x+2)=0 equation:

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

We get rid of parentheses

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

We multiply parentheses ..

(+3x^2+9x+6x+18)+x+2=0

We get rid of parentheses

3x^2+9x+6x+x+18+2=0

We add all the numbers together, and all the variables

3x^2+16x+20=0

a = 3; b = 16; c = +20;
Δ = b2-4ac
Δ = 162-4·3·20
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4}{2*3}=\frac{-20}{6}
=-3+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4}{2*3}=\frac{-12}{6}
=-2
$