k^2=-4k+12

Solution for k^2=-4k+12 equation:

k^2=-4k+12

We move all terms to the left:

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

We get rid of parentheses

k^2+4k-12=0

a = 1; b = 4; c = -12;
Δ = b2-4ac
Δ = 42-4·1·(-12)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*1}=\frac{-12}{2}
=-6
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*1}=\frac{4}{2}
=2
$