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k^2+10k-11=0
a = 1; b = 10; c = -11;
Δ = b2-4ac
Δ = 102-4·1·(-11)
Δ = 144
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{144}=12$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-12}{2*1}=\frac{-22}{2} =-11 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+12}{2*1}=\frac{2}{2} =1 $
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