(2x+1)(3x+3)=9

Solution for (2x+1)(3x+3)=9 equation:

(2x+1)(3x+3)=9

We move all terms to the left:

(2x+1)(3x+3)-(9)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2+9x-6=0

a = 6; b = 9; c = -6;
Δ = b2-4ac
Δ = 92-4·6·(-6)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-15}{2*6}=\frac{-24}{12}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*6}=\frac{6}{12}
=1/2
$