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z^2+11z-28=0
a = 1; b = 11; c = -28;
Δ = b2-4ac
Δ = 112-4·1·(-28)
Δ = 233
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{-(11)-\sqrt{233}}{2*1}=\frac{-11-\sqrt{233}}{2} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{233}}{2*1}=\frac{-11+\sqrt{233}}{2} $
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