p^2+7p-98=0

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Solution for p^2+7p-98=0 equation: 



p^2+7p-98=0

a = 1; b = 7; c = -98;
Δ = b2-4ac
Δ = 72-4·1·(-98)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$


$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-21}{2*1}=\frac{-28}{2}
=-14
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+21}{2*1}=\frac{14}{2}
=7
$

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