p^2+6p-91=0

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Solution for p^2+6p-91=0 equation: 



p^2+6p-91=0

a = 1; b = 6; c = -91;
Δ = b2-4ac
Δ = 62-4·1·(-91)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$


$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-20}{2*1}=\frac{-26}{2}
=-13
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+20}{2*1}=\frac{14}{2}
=7
$

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