p^2+14p+45=0

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Solution for p^2+14p+45=0 equation: 



p^2+14p+45=0

a = 1; b = 14; c = +45;
Δ = b2-4ac
Δ = 142-4·1·45
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4}{2*1}=\frac{-18}{2}
=-9
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4}{2*1}=\frac{-10}{2}
=-5
$

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