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

Solution for 6=(x-3)(2x-4) equation:

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

We move all terms to the left:

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

-2x^2+10x-6=0

a = -2; b = 10; c = -6;
Δ = b2-4ac
Δ = 102-4·(-2)·(-6)
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{13}}{2*-2}=\frac{-10-2\sqrt{13}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{13}}{2*-2}=\frac{-10+2\sqrt{13}}{-4}
$