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k^2+13k-14=0
a = 1; b = 13; c = -14;
Δ = b2-4ac
Δ = 132-4·1·(-14)
Δ = 225
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{225}=15$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-15}{2*1}=\frac{-28}{2} =-14 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+15}{2*1}=\frac{2}{2} =1 $
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