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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3A^2-20A+12

Back to the equation:

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


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3A^2+21A-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:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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