(3x-2)(x+4)=(3x-8)(2x-3)

Solution for (3x-2)(x+4)=(3x-8)(2x-3) equation:

(3x-2)(x+4)=(3x-8)(2x-3)

We move all terms to the left:

(3x-2)(x+4)-((3x-8)(2x-3))=0

We multiply parentheses ..

(+3x^2+12x-2x-8)-((3x-8)(2x-3))=0


We calculate terms in parentheses: -((3x-8)(2x-3)), so:

(3x-8)(2x-3)

We multiply parentheses ..

(+6x^2-9x-16x+24)

We get rid of parentheses

6x^2-9x-16x+24

We add all the numbers together, and all the variables

6x^2-25x+24

Back to the equation:

-(6x^2-25x+24)


We get rid of parentheses

3x^2-6x^2+12x-2x+25x-8-24=0

We add all the numbers together, and all the variables

-3x^2+35x-32=0

a = -3; b = 35; c = -32;
Δ = b2-4ac
Δ = 352-4·(-3)·(-32)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-29}{2*-3}=\frac{-64}{-6}
=10+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+29}{2*-3}=\frac{-6}{-6}
=1
$