2x^2+7x>= 0

Lösung

2 x^{2} + 7 x \geq 0

x \leq - \frac{7}{2} or x \geq 0

Intervall-Notation

(- \infty, - \frac{7}{2}] \cup [0, \infty)

Dezimale

x \leq - 3.5 or x \geq 0

Schritte zur Lösung

2 x^{2} + 7 x \geq 0

Faktorisiere

2 x^{2} + 7 x : x (2 x + 7)

x (2 x + 7) \geq 0

Identifiziere die Intervalle

x \leq - \frac{7}{2} or x \geq 0

x^{2} - 9 x + 38 = 0

8 m = 4

22 \leq 7 x + 1 < 50

4 w^{2} + 12 w + 9 = 121

3 x + 1 = 5 x - 9