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(P)=2P^2=5P-1
We move all terms to the left:
(P)-(2P^2)=0
determiningTheFunctionDomain -2P^2+P=0
a = -2; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-2)·0
Δ = 1
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{1}=1$$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-2}=\frac{-2}{-4} =1/2 $$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-2}=\frac{0}{-4} =0 $
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