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k^2+19k+18=0
a = 1; b = 19; c = +18;
Δ = b2-4ac
Δ = 192-4·1·18
Δ = 289
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{289}=17$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-17}{2*1}=\frac{-36}{2} =-18 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+17}{2*1}=\frac{-2}{2} =-1 $
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