p^2+3p-1=0

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Solution for p^2+3p-1=0 equation: 



p^2+3p-1=0

a = 1; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·1·(-1)
Δ = 13
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{13}}{2*1}=\frac{-3-\sqrt{13}}{2}
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{13}}{2*1}=\frac{-3+\sqrt{13}}{2}
$

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