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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -6; b = -12; c = +48;
Δ = b2-4ac
Δ = -122-4·(-6)·48
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-36}{2*-6}=\frac{-24}{-12}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+36}{2*-6}=\frac{48}{-12}
=-4
$