p^2+7p-4=0

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Solution for p^2+7p-4=0 equation: 



p^2+7p-4=0

a = 1; b = 7; c = -4;
Δ = b2-4ac
Δ = 72-4·1·(-4)
Δ = 65
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{-(7)-\sqrt{65}}{2*1}=\frac{-7-\sqrt{65}}{2}
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{65}}{2*1}=\frac{-7+\sqrt{65}}{2}
$

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