k2=196

Solution for k2=196 equation:

k2=196

We move all terms to the left:

k2-(196)=0

We add all the numbers together, and all the variables

k^2-196=0

a = 1; b = 0; c = -196;
Δ = b2-4ac
Δ = 02-4·1·(-196)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28}{2*1}=\frac{-28}{2}
=-14
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28}{2*1}=\frac{28}{2}
=14
$