2k^2-4K-23=0

Solution for 2k^2-4K-23=0 equation:

2k^2-4-23=0

We add all the numbers together, and all the variables

2k^2-27=0

a = 2; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·2·(-27)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{6}}{2*2}=\frac{0-6\sqrt{6}}{4}
=-\frac{6\sqrt{6}}{4}
=-\frac{3\sqrt{6}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{6}}{2*2}=\frac{0+6\sqrt{6}}{4}
=\frac{6\sqrt{6}}{4}
=\frac{3\sqrt{6}}{2}
$