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

Solution for (3x-12)(2x-4)=0 equation:

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

We multiply parentheses ..

(+6x^2-12x-24x+48)=0

We get rid of parentheses

6x^2-12x-24x+48=0

We add all the numbers together, and all the variables

6x^2-36x+48=0

a = 6; b = -36; c = +48;
Δ = b2-4ac
Δ = -362-4·6·48
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-12}{2*6}=\frac{24}{12}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+12}{2*6}=\frac{48}{12}
=4
$