(x+1)(3x+2)=(5x+-4)(3x+-2)

Solution for (x+1)(3x+2)=(5x+-4)(3x+-2) equation:

(x+1)(3x+2)=(5x+-4)(3x+-2)

We move all terms to the left:

(x+1)(3x+2)-((5x+-4)(3x+-2))=0

We add all the numbers together, and all the variables

(x+1)(3x+2)-((5x-4)(3x-2))=0

We multiply parentheses ..

(+3x^2+2x+3x+2)-((5x-4)(3x-2))=0


We calculate terms in parentheses: -((5x-4)(3x-2)), so:

(5x-4)(3x-2)

We multiply parentheses ..

(+15x^2-10x-12x+8)

We get rid of parentheses

15x^2-10x-12x+8

We add all the numbers together, and all the variables

15x^2-22x+8

Back to the equation:

-(15x^2-22x+8)


We get rid of parentheses

3x^2-15x^2+2x+3x+22x+2-8=0

We add all the numbers together, and all the variables

-12x^2+27x-6=0

a = -12; b = 27; c = -6;
Δ = b2-4ac
Δ = 272-4·(-12)·(-6)
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{-(27)-21}{2*-12}=\frac{-48}{-24}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+21}{2*-12}=\frac{-6}{-24}
=1/4
$