(2x+1)(3x+2)-(x-3)(x+2)=3(x+1)(3x-4)

Solution for (2x+1)(3x+2)-(x-3)(x+2)=3(x+1)(3x-4) equation:

(2x+1)(3x+2)-(x-3)(x+2)=3(x+1)(3x-4)

We move all terms to the left:

(2x+1)(3x+2)-(x-3)(x+2)-(3(x+1)(3x-4))=0

We multiply parentheses ..

(+6x^2+4x+3x+2)-(x-3)(x+2)-(3(x+1)(3x-4))=0


We calculate terms in parentheses: -(3(x+1)(3x-4)), so:

3(x+1)(3x-4)

We multiply parentheses ..

3(+3x^2-4x+3x-4)

We multiply parentheses

9x^2-12x+9x-12

We add all the numbers together, and all the variables

9x^2-3x-12

Back to the equation:

-(9x^2-3x-12)


We get rid of parentheses

6x^2-9x^2+4x+3x-(x-3)(x+2)+3x+2+12=0

We multiply parentheses ..

6x^2-9x^2-(+x^2+2x-3x-6)+4x+3x+3x+2+12=0

We add all the numbers together, and all the variables

-3x^2-(+x^2+2x-3x-6)+10x+14=0

We get rid of parentheses

-3x^2-x^2-2x+3x+10x+6+14=0

We add all the numbers together, and all the variables

-4x^2+11x+20=0

a = -4; b = 11; c = +20;
Δ = b2-4ac
Δ = 112-4·(-4)·20
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-21}{2*-4}=\frac{-32}{-8}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+21}{2*-4}=\frac{10}{-8}
=-1+1/4
$