(2x-8)(3x+1)=(4x-12)(x-2)+8

Solution for (2x-8)(3x+1)=(4x-12)(x-2)+8 equation:

(2x-8)(3x+1)=(4x-12)(x-2)+8

We move all terms to the left:

(2x-8)(3x+1)-((4x-12)(x-2)+8)=0

We multiply parentheses ..

(+6x^2+2x-24x-8)-((4x-12)(x-2)+8)=0


We calculate terms in parentheses: -((4x-12)(x-2)+8), so:

(4x-12)(x-2)+8

We multiply parentheses ..

(+4x^2-8x-12x+24)+8

We get rid of parentheses

4x^2-8x-12x+24+8

We add all the numbers together, and all the variables

4x^2-20x+32

Back to the equation:

-(4x^2-20x+32)


We get rid of parentheses

6x^2-4x^2+2x-24x+20x-8-32=0

We add all the numbers together, and all the variables

2x^2-2x-40=0

a = 2; b = -2; c = -40;
Δ = b2-4ac
Δ = -22-4·2·(-40)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-18}{2*2}=\frac{-16}{4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+18}{2*2}=\frac{20}{4}
=5
$