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z^2+3z-270=0
a = 1; b = 3; c = -270;
Δ = b2-4ac
Δ = 32-4·1·(-270)
Δ = 1089
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{1089}=33$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-33}{2*1}=\frac{-36}{2} =-18 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+33}{2*1}=\frac{30}{2} =15 $
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