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k2=36
We move all terms to the left:
k2-(36)=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 $
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