7p^2+7p-3=0

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Solution for 7p^2+7p-3=0 equation: 



7p^2+7p-3=0

a = 7; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·7·(-3)
Δ = 133
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{133}}{2*7}=\frac{-7-\sqrt{133}}{14}
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{133}}{2*7}=\frac{-7+\sqrt{133}}{14}
$

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