3(k-8)(k+2)=0

Solution for 3(k-8)(k+2)=0 equation:

3(k-8)(k+2)=0

We multiply parentheses ..

3(+k^2+2k-8k-16)=0

We multiply parentheses

3k^2+6k-24k-48=0

We add all the numbers together, and all the variables

3k^2-18k-48=0

a = 3; b = -18; c = -48;
Δ = b2-4ac
Δ = -182-4·3·(-48)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-30}{2*3}=\frac{-12}{6}
=-2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+30}{2*3}=\frac{48}{6}
=8
$