(2k)^2-4(2)(3k)=0

Solution for (2k)^2-4(2)(3k)=0 equation:

(2k)^2-4(2)(3k)=0

a = 2; b = -423; c = 0;
Δ = b2-4ac
Δ = -4232-4·2·0
Δ = 178929
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{178929}=423$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-423)-423}{2*2}=\frac{0}{4}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-423)+423}{2*2}=\frac{846}{4}
=211+1/2
$