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k^2+36k-76=0
a = 1; b = 36; c = -76;
Δ = b2-4ac
Δ = 362-4·1·(-76)
Δ = 1600
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{1600}=40$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-40}{2*1}=\frac{-76}{2} =-38 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+40}{2*1}=\frac{4}{2} =2 $
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