4k(k-2)=12

Solution for 4k(k-2)=12 equation:

4k(k-2)=12

We move all terms to the left:

4k(k-2)-(12)=0

We multiply parentheses

4k^2-8k-12=0

a = 4; b = -8; c = -12;
Δ = b2-4ac
Δ = -82-4·4·(-12)
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*4}=\frac{-8}{8}
=-1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*4}=\frac{24}{8}
=3
$