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

Solution for 0=-(2x-3)(4x-2) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -(-(+8x^2-4x-12x+6)), so:

-(+8x^2-4x-12x+6)

We get rid of parentheses

-8x^2+4x+12x-6

We add all the numbers together, and all the variables

-8x^2+16x-6

Back to the equation:

-(-8x^2+16x-6)


We get rid of parentheses

8x^2-16x+6=0

a = 8; b = -16; c = +6;
Δ = b2-4ac
Δ = -162-4·8·6
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8}{2*8}=\frac{8}{16}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8}{2*8}=\frac{24}{16}
=1+1/2
$