k^2=26

Solution for k^2=26 equation:

k^2=26

We move all terms to the left:

k^2-(26)=0

a = 1; b = 0; c = -26;
Δ = b2-4ac
Δ = 02-4·1·(-26)
Δ = 104
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:

$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{26}}{2*1}=\frac{0-2\sqrt{26}}{2}
=-\frac{2\sqrt{26}}{2}
=-\sqrt{26}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{26}}{2*1}=\frac{0+2\sqrt{26}}{2}
=\frac{2\sqrt{26}}{2}
=\sqrt{26}
$