k^2+6=209

Solution for k^2+6=209 equation:

k^2+6=209

We move all terms to the left:

k^2+6-(209)=0

We add all the numbers together, and all the variables

k^2-203=0

a = 1; b = 0; c = -203;
Δ = b2-4ac
Δ = 02-4·1·(-203)
Δ = 812
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:

$\sqrt{\Delta}=\sqrt{812}=\sqrt{4*203}=\sqrt{4}*\sqrt{203}=2\sqrt{203}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{203}}{2*1}=\frac{0-2\sqrt{203}}{2}
=-\frac{2\sqrt{203}}{2}
=-\sqrt{203}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{203}}{2*1}=\frac{0+2\sqrt{203}}{2}
=\frac{2\sqrt{203}}{2}
=\sqrt{203}
$