0.112=0.0418+0.0644x-0.00141x^2

Lösung

0.112 = 0.0418 + 0.0644 x - 0.00141 x^{2}

x = \frac{0.0644 - 0.003751432}{0.00282}, x = \frac{0.0644 + 0.003751432}{0.00282}

Dezimale

x = 1.11739 \dots, x = 44.55635 \dots

Schritte zur Lösung

0.112 = 0.0418 + 0.0644 x - 0.00141 x^{2}

Tausche die Seiten

0.0418 + 0.0644 x - 0.00141 x^{2} = 0.112

Verschiebe

0.112

auf die linke Seite

- 0.00141 x^{2} + 0.0644 x - 0.0702 = 0

Löse mit der quadratischen Formel

x_{1, 2} = \frac{- 0.0644 \pm 0.064 4 ^{2} - 4 ( - 0.00141 ) ( - 0.0702 )}{2 ( - 0.00141 )}

0.064 4^{2} - 4 (- 0.00141) (- 0.0702) = 0.003751432

x_{1, 2} = \frac{- 0.0644 \pm 0.003751432}{2 ( - 0.00141 )}

Trenne die Lösungen

x_{1} = \frac{- 0.0644 + 0.003751432}{2 ( - 0.00141 )}, x_{2} = \frac{- 0.0644 - 0.003751432}{2 ( - 0.00141 )}

x = \frac{- 0.0644 + 0.003751432}{2 ( - 0.00141 )} : \frac{0.0644 - 0.003751432}{0.00282}

x = \frac{- 0.0644 - 0.003751432}{2 ( - 0.00141 )} : \frac{0.0644 + 0.003751432}{0.00282}

Die Lösungen für die quadratische Gleichung sind:

x = \frac{0.0644 - 0.003751432}{0.00282}, x = \frac{0.0644 + 0.003751432}{0.00282}

- 1 < \frac{r}{12}

\frac{6}{x - 4} \geq \frac{3}{x - 3}

55 = 44 + x

f a c t or - 28 x^{7} y^{2} + 4 x^{4} y^{6}

∣ 1 - 3 x ∣ + 2 \geq 5