(2m-1)x^2+mx+15=0

Lösung

(2 m - 1) x^{2} + m x + 15 = 0

x = \frac{- m + m ^{2} - 60 ( 2 m - 1 )}{2 ( 2 m - 1 )}, x = \frac{- m - m ^{2} - 60 ( 2 m - 1 )}{2 ( 2 m - 1 )}; m \neq = \frac{1}{2}

Schritte zur Lösung

(2 m - 1) x^{2} + m x + 15 = 0

Löse mit der quadratischen Formel

x_{1, 2} = \frac{- m \pm m ^{2} - 4 ( 2 m - 1 ) \cdot 15}{2 ( 2 m - 1 )}

Vereinfache

m^{2} - 4 (2 m - 1) \cdot 15 : m^{2} - 60 (2 m - 1)

x_{1, 2} = \frac{- m \pm m ^{2} - 60 ( 2 m - 1 )}{2 ( 2 m - 1 )}; m \neq = \frac{1}{2}

Trenne die Lösungen

x_{1} = \frac{- m + m ^{2} - 60 ( 2 m - 1 )}{2 ( 2 m - 1 )}, x_{2} = \frac{- m - m ^{2} - 60 ( 2 m - 1 )}{2 ( 2 m - 1 )}

x = \frac{- m + m ^{2} - 60 ( 2 m - 1 )}{2 ( 2 m - 1 )}; m \neq = \frac{1}{2}

x = \frac{- m - m ^{2} - 60 ( 2 m - 1 )}{2 ( 2 m - 1 )}; m \neq = \frac{1}{2}

Die Lösungen für die quadratische Gleichung sind:

x = \frac{- m + m ^{2} - 60 ( 2 m - 1 )}{2 ( 2 m - 1 )}, x = \frac{- m - m ^{2} - 60 ( 2 m - 1 )}{2 ( 2 m - 1 )}; m \neq = \frac{1}{2}

2 - x (x + 4) = - 1 (x - 2)

v^{2} - v - 2 = 0

so l v e f or a, a^{2} + ab - 2 b^{2} = 0

13 + x^{2} = 10 x

6 πh x + 42 π x^{2} = 2160 π