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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

k^2+8k-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:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*1}=\frac{-24}{2}
=-12
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*1}=\frac{8}{2}
=4
$