2p^2+p-8=0

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Solution for 2p^2+p-8=0 equation: 



2p^2+p-8=0

a = 2; b = 1; c = -8;
Δ = b2-4ac
Δ = 12-4·2·(-8)
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{-(1)-\sqrt{65}}{2*2}=\frac{-1-\sqrt{65}}{4}
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{65}}{2*2}=\frac{-1+\sqrt{65}}{4}
$

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