0.5p^2+p-1=0

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Solution for 0.5p^2+p-1=0 equation: 



0.5p^2+p-1=0

a = 0.5; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·0.5·(-1)
Δ = 3
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{3}}{2*0.5}=\frac{-1-\sqrt{3}}{1}
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{3}}{2*0.5}=\frac{-1+\sqrt{3}}{1}
$

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