k2+6=19

Solution for k2+6=19 equation:

k2+6=19

We move all terms to the left:

k2+6-(19)=0

We add all the numbers together, and all the variables

k^2-13=0

a = 1; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·1·(-13)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{13}}{2*1}=\frac{0-2\sqrt{13}}{2}
=-\frac{2\sqrt{13}}{2}
=-\sqrt{13}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{13}}{2*1}=\frac{0+2\sqrt{13}}{2}
=\frac{2\sqrt{13}}{2}
=\sqrt{13}
$