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p^2+18p-88=0
a = 1; b = 18; c = -88;
Δ = b2-4ac
Δ = 182-4·1·(-88)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{676}=26$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-26}{2*1}=\frac{-44}{2} =-22 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+26}{2*1}=\frac{8}{2} =4 $
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