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k^2+13k-54=0
a = 1; b = 13; c = -54;
Δ = b2-4ac
Δ = 132-4·1·(-54)
Δ = 385
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{-(13)-\sqrt{385}}{2*1}=\frac{-13-\sqrt{385}}{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{385}}{2*1}=\frac{-13+\sqrt{385}}{2} $
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