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k^2-11k-24=0
a = 1; b = -11; c = -24;
Δ = b2-4ac
Δ = -112-4·1·(-24)
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{-(-11)-\sqrt{217}}{2*1}=\frac{11-\sqrt{217}}{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{217}}{2*1}=\frac{11+\sqrt{217}}{2} $
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