It is analogous to Marcus equation- where L is lembda to calculate
∆G^‡=L/4 [1-(∆G^0)/L]^2
Nicholas N Olenev
Popular answer
Dear researcher, your equation is solved elementarily if you change Y = 1 / X:
P^2 Y^2 - 2(P+2Q)Y + 1 = 0, Y_1,2 = (P + 2Q +- 2 (Q(P+Q))^(1/2)) / (P^2).
So,
X_1 = (P^2) / (P + 2Q + 2 (Q(P+Q))^(1/2)),
X_2 = (P^2) / (P + 2Q - 2 (Q(P+Q))^(1/2))
3 votes
1 thanks
Nicholas N Olenev
Dear researcher, your equation is solved elementarily if you change Y = 1 / X:
P^2 Y^2 - 2(P+2Q)Y + 1 = 0, Y_1,2 = (P + 2Q +- 2 (Q(P+Q))^(1/2)) / (P^2).
So,
X_1 = (P^2) / (P + 2Q + 2 (Q(P+Q))^(1/2)),
X_2 = (P^2) / (P + 2Q - 2 (Q(P+Q))^(1/2))
3 votes
3 thanks
Bhowmik Subrata
Better use MATHEMATICA software, input the equation and you will get accurate answer.
0 votes
0 thanks
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