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k2=6k=
We move all terms to the left:
k2-(6k)=0
We add all the numbers together, and all the variables
k^2-6k=0
a = 1; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·1·0
Δ = 36
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{36}=6$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*1}=\frac{0}{2} =0 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*1}=\frac{12}{2} =6 $
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