k^2=169

Solution for k^2=169 equation:

k^2=169

We move all terms to the left:

k^2-(169)=0

a = 1; b = 0; c = -169;
Δ = b2-4ac
Δ = 02-4·1·(-169)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-26}{2*1}=\frac{-26}{2}
=-13
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+26}{2*1}=\frac{26}{2}
=13
$