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(P)=P^2-4P+3
We move all terms to the left:
(P)-(P^2-4P+3)=0
We get rid of parentheses
-P^2+P+4P-3=0
We add all the numbers together, and all the variables
-1P^2+5P-3=0
a = -1; b = 5; c = -3;
Δ = b2-4ac
Δ = 52-4·(-1)·(-3)
Δ = 13
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{-(5)-\sqrt{13}}{2*-1}=\frac{-5-\sqrt{13}}{-2} $$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{13}}{2*-1}=\frac{-5+\sqrt{13}}{-2} $
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