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z^2-18z-19=0
a = 1; b = -18; c = -19;
Δ = b2-4ac
Δ = -182-4·1·(-19)
Δ = 400
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{400}=20$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-20}{2*1}=\frac{-2}{2} =-1 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+20}{2*1}=\frac{38}{2} =19 $
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