2k^2+4=28

Solution for 2k^2+4=28 equation:

2k^2+4=28

We move all terms to the left:

2k^2+4-(28)=0

We add all the numbers together, and all the variables

2k^2-24=0

a = 2; b = 0; c = -24;
Δ = b2-4ac
Δ = 02-4·2·(-24)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*2}=\frac{0-8\sqrt{3}}{4}
=-\frac{8\sqrt{3}}{4}
=-2\sqrt{3}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*2}=\frac{0+8\sqrt{3}}{4}
=\frac{8\sqrt{3}}{4}
=2\sqrt{3}
$