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k^2-7k-5=0
a = 1; b = -7; c = -5;
Δ = b2-4ac
Δ = -72-4·1·(-5)
Δ = 69
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{-(-7)-\sqrt{69}}{2*1}=\frac{7-\sqrt{69}}{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{69}}{2*1}=\frac{7+\sqrt{69}}{2} $
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