(3x-8)(2x+4)=(4x+8)(3x-5)

Solution for (3x-8)(2x+4)=(4x+8)(3x-5) equation:

(3x-8)(2x+4)=(4x+8)(3x-5)

We move all terms to the left:

(3x-8)(2x+4)-((4x+8)(3x-5))=0

We multiply parentheses ..

(+6x^2+12x-16x-32)-((4x+8)(3x-5))=0


We calculate terms in parentheses: -((4x+8)(3x-5)), so:

(4x+8)(3x-5)

We multiply parentheses ..

(+12x^2-20x+24x-40)

We get rid of parentheses

12x^2-20x+24x-40

We add all the numbers together, and all the variables

12x^2+4x-40

Back to the equation:

-(12x^2+4x-40)


We get rid of parentheses

6x^2-12x^2+12x-16x-4x-32+40=0

We add all the numbers together, and all the variables

-6x^2-8x+8=0

a = -6; b = -8; c = +8;
Δ = b2-4ac
Δ = -82-4·(-6)·8
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16}{2*-6}=\frac{-8}{-12}
=2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*-6}=\frac{24}{-12}
=-2
$