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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+x^2-12x+4x-48)=0

We get rid of parentheses

x^2-12x+4x-48=0

We add all the numbers together, and all the variables

x^2-8x-48=0

a = 1; b = -8; c = -48;
Δ = b2-4ac
Δ = -82-4·1·(-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{-(-8)-16}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*1}=\frac{24}{2}
=12
$