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

Solution for (x+2)(x+3)=(x-3)(x-2)-2x(x+1)=0 equation:

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((x-3)(x-2)-2x(x+1)), so:

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

We multiply parentheses

-2x^2+(x-3)(x-2)-2x

We multiply parentheses ..

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

We get rid of parentheses

-2x^2+x^2-2x-3x-2x+6

We add all the numbers together, and all the variables

-1x^2-7x+6

Back to the equation:

-(-1x^2-7x+6)


We get rid of parentheses

x^2+1x^2+3x+2x+7x+6-6=0

We add all the numbers together, and all the variables

2x^2+12x=0

a = 2; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·2·0
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{-(12)-12}{2*2}=\frac{-24}{4}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*2}=\frac{0}{4}
=0
$