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racx+3x-2 < 4, \quad x - Anadea

📅 May 15, 2026 👤 scraface

May 15, 2026

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2(x+1) + 3x = x(x+1)
2x + 2 + 3x = x^2 + x
5x + 2 = x^2 + x
Bringe alle Terme auf eine Seite:
= x^2 + x - 5x - 2
x^2 - 4x - 2 = 0
Verwende die quadratische Lösungsformel:
x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-2)}}{2(1)} = \frac{4 \pm \sqrt{16...
Somit sind die Lösungen \( x = 2 + \sqrt{6} \) und \( x = 2 - \sqrt{6}...
\(\boxed{2 \pm \sqrt{6}}\)1. **Frage:

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