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2p+1=p^2
We move all terms to the left:
2p+1-(p^2)=0
determiningTheFunctionDomain -p^2+2p+1=0
We add all the numbers together, and all the variables
-1p^2+2p+1=0
a = -1; b = 2; c = +1;
Δ = b2-4ac
Δ = 22-4·(-1)·1
Δ = 8
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{2}}{2*-1}=\frac{-2-2\sqrt{2}}{-2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{2}}{2*-1}=\frac{-2+2\sqrt{2}}{-2} $
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