-2p^2+p+2=0

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



-2p^2+p+2=0

a = -2; b = 1; c = +2;
Δ = b2-4ac
Δ = 12-4·(-2)·2
Δ = 17
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{17}}{2*-2}=\frac{-1-\sqrt{17}}{-4}
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{17}}{2*-2}=\frac{-1+\sqrt{17}}{-4}
$

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