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k^2-20k-800=0
a = 1; b = -20; c = -800;
Δ = b2-4ac
Δ = -202-4·1·(-800)
Δ = 3600
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{3600}=60$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-60}{2*1}=\frac{-40}{2} =-20 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+60}{2*1}=\frac{80}{2} =40 $
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