k2+6=23

Solution for k2+6=23 equation:

k2+6=23

We move all terms to the left:

k2+6-(23)=0

We add all the numbers together, and all the variables

k^2-17=0

a = 1; b = 0; c = -17;
Δ = b2-4ac
Δ = 02-4·1·(-17)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{17}}{2*1}=\frac{0-2\sqrt{17}}{2}
=-\frac{2\sqrt{17}}{2}
=-\sqrt{17}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{17}}{2*1}=\frac{0+2\sqrt{17}}{2}
=\frac{2\sqrt{17}}{2}
=\sqrt{17}
$