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k^2-8k-5=0
a = 1; b = -8; c = -5;
Δ = b2-4ac
Δ = -82-4·1·(-5)
Δ = 84
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{21}}{2*1}=\frac{8-2\sqrt{21}}{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{21}}{2*1}=\frac{8+2\sqrt{21}}{2} $
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