12=(10-2x)(6-2x)

Solution for 12=(10-2x)(6-2x) equation:

12=(10-2x)(6-2x)

We move all terms to the left:

12-((10-2x)(6-2x))=0

We add all the numbers together, and all the variables

-((-2x+10)(-2x+6))+12=0

We multiply parentheses ..

-((+4x^2-12x-20x+60))+12=0


We calculate terms in parentheses: -((+4x^2-12x-20x+60)), so:

(+4x^2-12x-20x+60)

We get rid of parentheses

4x^2-12x-20x+60

We add all the numbers together, and all the variables

4x^2-32x+60

Back to the equation:

-(4x^2-32x+60)


We get rid of parentheses

-4x^2+32x-60+12=0

We add all the numbers together, and all the variables

-4x^2+32x-48=0

a = -4; b = 32; c = -48;
Δ = b2-4ac
Δ = 322-4·(-4)·(-48)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16}{2*-4}=\frac{-48}{-8}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16}{2*-4}=\frac{-16}{-8}
=+2
$