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-1+p=-2p^2
We move all terms to the left:
-1+p-(-2p^2)=0
We get rid of parentheses
2p^2+p-1=0
a = 2; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·2·(-1)
Δ = 9
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9}=3$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3}{2*2}=\frac{-4}{4} =-1 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3}{2*2}=\frac{2}{4} =1/2 $
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