2k^2-12k+8=0

Solution for 2k^2-12k+8=0 equation:

2k^2-12k+8=0

a = 2; b = -12; c = +8;
Δ = b2-4ac
Δ = -122-4·2·8
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{5}}{2*2}=\frac{12-4\sqrt{5}}{4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{5}}{2*2}=\frac{12+4\sqrt{5}}{4}
$