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k2-5k=1
We move all terms to the left:
k2-5k-(1)=0
We add all the numbers together, and all the variables
k^2-5k-1=0
a = 1; b = -5; c = -1;
Δ = b2-4ac
Δ = -52-4·1·(-1)
Δ = 29
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{29}}{2*1}=\frac{5-\sqrt{29}}{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{29}}{2*1}=\frac{5+\sqrt{29}}{2} $
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