(2x-4)(-12x+8)=4(1x-2)(10x-2)

Solution for (2x-4)(-12x+8)=4(1x-2)(10x-2) equation:

(2x-4)(-12x+8)=4(1x-2)(10x-2)

We move all terms to the left:

(2x-4)(-12x+8)-(4(1x-2)(10x-2))=0

We add all the numbers together, and all the variables

(2x-4)(-12x+8)-(4(x-2)(10x-2))=0

We multiply parentheses ..

(-24x^2+16x+48x-32)-(4(x-2)(10x-2))=0


We calculate terms in parentheses: -(4(x-2)(10x-2)), so:

4(x-2)(10x-2)

We multiply parentheses ..

4(+10x^2-2x-20x+4)

We multiply parentheses

40x^2-8x-80x+16

We add all the numbers together, and all the variables

40x^2-88x+16

Back to the equation:

-(40x^2-88x+16)


We get rid of parentheses

-24x^2-40x^2+16x+48x+88x-32-16=0

We add all the numbers together, and all the variables

-64x^2+152x-48=0

a = -64; b = 152; c = -48;
Δ = b2-4ac
Δ = 1522-4·(-64)·(-48)
Δ = 10816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10816}=104$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(152)-104}{2*-64}=\frac{-256}{-128}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(152)+104}{2*-64}=\frac{-48}{-128}
=3/8
$