k2+5=41

Solution for k2+5=41 equation:

k2+5=41

We move all terms to the left:

k2+5-(41)=0

We add all the numbers together, and all the variables

k^2-36=0

a = 1; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·1·(-36)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*1}=\frac{-12}{2}
=-6
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*1}=\frac{12}{2}
=6
$