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z^2-5z-45=0
a = 1; b = -5; c = -45;
Δ = b2-4ac
Δ = -52-4·1·(-45)
Δ = 205
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{205}}{2*1}=\frac{5-\sqrt{205}}{2} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{205}}{2*1}=\frac{5+\sqrt{205}}{2} $
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