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z^2-10z-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:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-12}{2*1}=\frac{-2}{2} =-1 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+12}{2*1}=\frac{22}{2} =11 $
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