(2m-3)x^2+(m-3)x-2m+1=0

Lösung

(2 m - 3) x^{2} + (m - 3) x - 2 m + 1 = 0

x = \frac{- m + 3 + 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}, x = \frac{- m + 3 - 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}; m \neq = \frac{3}{2}

Schritte zur Lösung

(2 m - 3) x^{2} + (m - 3) x - 2 m + 1 = 0

Löse mit der quadratischen Formel

x_{1, 2} = \frac{- ( m - 3 ) \pm ( m - 3 ) ^{2} - 4 ( 2 m - 3 ) ( - 2 m + 1 )}{2 ( 2 m - 3 )}

Vereinfache

(m - 3)^{2} - 4 (2 m - 3) (- 2 m + 1) : 17 m^{2} - 38 m + 21

x_{1, 2} = \frac{- ( m - 3 ) \pm 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}; m \neq = \frac{3}{2}

Trenne die Lösungen

x_{1} = \frac{- ( m - 3 ) + 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}, x_{2} = \frac{- ( m - 3 ) - 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}

x = \frac{- ( m - 3 ) + 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )} : \frac{- m + 3 + 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}

x = \frac{- ( m - 3 ) - 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )} : \frac{- m + 3 - 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}

Die Lösungen für die quadratische Gleichung sind:

x = \frac{- m + 3 + 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}, x = \frac{- m + 3 - 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}; m \neq = \frac{3}{2}

p^{2} - 100 = 0

- 28 q = - 96 q^{2} - 2

- 3 x^{2} - 30 = 0

so l v e f or x, e^{y} = x^{2} + y^{2}

x^{2} + 40 x + 400 = 0