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p^2+18p+65=0
a = 1; b = 18; c = +65;
Δ = b2-4ac
Δ = 182-4·1·65
Δ = 64
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{64}=8$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-8}{2*1}=\frac{-26}{2} =-13 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+8}{2*1}=\frac{-10}{2} =-5 $
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