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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+6x^2-15x-4x+10)-(3x-2)(x+1))+x=0


We calculate terms in parentheses: -((+6x^2-15x-4x+10)-(3x-2)(x+1)), so:

(+6x^2-15x-4x+10)-(3x-2)(x+1)

We get rid of parentheses

6x^2-15x-4x-(3x-2)(x+1)+10

We multiply parentheses ..

6x^2-(+3x^2+3x-2x-2)-15x-4x+10

We add all the numbers together, and all the variables

6x^2-(+3x^2+3x-2x-2)-19x+10

We get rid of parentheses

6x^2-3x^2-3x+2x-19x+2+10

We add all the numbers together, and all the variables

3x^2-20x+12

Back to the equation:

-(3x^2-20x+12)


We add all the numbers together, and all the variables

x-(3x^2-20x+12)=0

We get rid of parentheses

-3x^2+x+20x-12=0

We add all the numbers together, and all the variables

-3x^2+21x-12=0

a = -3; b = 21; c = -12;
Δ = b2-4ac
Δ = 212-4·(-3)·(-12)
Δ = 297
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-3\sqrt{33}}{2*-3}=\frac{-21-3\sqrt{33}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+3\sqrt{33}}{2*-3}=\frac{-21+3\sqrt{33}}{-6}
$