12=(x-3)(2x-4)=12

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

12=(x-3)(2x-4)=12

We move all terms to the left:

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

We multiply parentheses ..

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


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

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-10x+12

Back to the equation:

-(2x^2-10x+12)


We get rid of parentheses

-2x^2+10x-12+12=0

We add all the numbers together, and all the variables

-2x^2+10x=0

a = -2; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-2)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*-2}=\frac{-20}{-4}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*-2}=\frac{0}{-4}
=0
$