x2 + (-3b + 2,5c)·x - 7,5bc = 0
x2 + px + q = 0
pq-Formel: p = -3b + 2,5c ; q = -15b
→ p/2 = -1,5b + 1,25c → (p/2)2 = (-3/2b + 5/4c)2 = 9·b2/4 - 15·b·c/4 + 25·c2/16
x1,2 = - p/2 ± \(\sqrt{(p/2)^2 - q}\)
x1,2 = 1,5b - 1,25c ± \(\sqrt{ 9·b^2/4 - 15·b·c/4 + 25·c^2/16+7,5bc}\)
x1,2 = 1,5b - 1,25c ± \(\sqrt{ (36b^2 + 60bc + 25c^2) / 16 }\)
unter der Wurzel 1. binomische Formel anwenden:
x1,2 = 1,5b - 1,25c ± \(\sqrt{ (6b +5c)^2 /16 }\)
x1,2 = 1,5b - 1,25c ± (6b +5c) / 4
x1,2 = 1,5b -1,25c ± 1,5b ± 1,25c
x1 = 3b ; x2 = - 2,5c
Gruß Wolfgang