k2+6=18

Solution for k2+6=18 equation:

k2+6=18

We move all terms to the left:

k2+6-(18)=0

We add all the numbers together, and all the variables

k^2-12=0

a = 1; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·1·(-12)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*1}=\frac{0-4\sqrt{3}}{2}
=-\frac{4\sqrt{3}}{2}
=-2\sqrt{3}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*1}=\frac{0+4\sqrt{3}}{2}
=\frac{4\sqrt{3}}{2}
=2\sqrt{3}
$