(k-8)(k+4)=0

Solution for (k-8)(k+4)=0 equation:

(k-8)(k+4)=0

We multiply parentheses ..

(+k^2+4k-8k-32)=0

We get rid of parentheses

k^2+4k-8k-32=0

We add all the numbers together, and all the variables

k^2-4k-32=0

a = 1; b = -4; c = -32;
Δ = b2-4ac
Δ = -42-4·1·(-32)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*1}=\frac{-8}{2}
=-4
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*1}=\frac{16}{2}
=8
$