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z^2+15z-20=0
a = 1; b = 15; c = -20;
Δ = b2-4ac
Δ = 152-4·1·(-20)
Δ = 305
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{-(15)-\sqrt{305}}{2*1}=\frac{-15-\sqrt{305}}{2} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{305}}{2*1}=\frac{-15+\sqrt{305}}{2} $
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