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k^2-13k-12=0
a = 1; b = -13; c = -12;
Δ = b2-4ac
Δ = -132-4·1·(-12)
Δ = 217
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{217}}{2*1}=\frac{13-\sqrt{217}}{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{217}}{2*1}=\frac{13+\sqrt{217}}{2} $
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