k^2-12k+4=0

Solution for k^2-12k+4=0 equation:

k^2-12k+4=0

a = 1; b = -12; c = +4;
Δ = b2-4ac
Δ = -122-4·1·4
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{2}}{2*1}=\frac{12-8\sqrt{2}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{2}}{2*1}=\frac{12+8\sqrt{2}}{2}
$