7p^2+p-4=0

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



7p^2+p-4=0

a = 7; b = 1; c = -4;
Δ = b2-4ac
Δ = 12-4·7·(-4)
Δ = 113
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{-(1)-\sqrt{113}}{2*7}=\frac{-1-\sqrt{113}}{14}
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{113}}{2*7}=\frac{-1+\sqrt{113}}{14}
$

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