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k^2+19k+33=0
a = 1; b = 19; c = +33;
Δ = b2-4ac
Δ = 192-4·1·33
Δ = 229
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{-(19)-\sqrt{229}}{2*1}=\frac{-19-\sqrt{229}}{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{229}}{2*1}=\frac{-19+\sqrt{229}}{2} $
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