p^2+18p+65=0

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Solution for p^2+18p+65=0 equation:



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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