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k^2+20k-44=0
a = 1; b = 20; c = -44;
Δ = b2-4ac
Δ = 202-4·1·(-44)
Δ = 576
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{576}=24$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-24}{2*1}=\frac{-44}{2} =-22 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+24}{2*1}=\frac{4}{2} =2 $
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