x^2-8x+15>= 0

Lösung

x^{2} - 8 x + 15 \geq 0

x \leq 3 or x \geq 5

Intervall-Notation

(- \infty, 3] \cup [5, \infty)

Schritte zur Lösung

x^{2} - 8 x + 15 \geq 0

Faktorisiere

x^{2} - 8 x + 15 : (x - 3) (x - 5)

(x - 3) (x - 5) \geq 0

Identifiziere die Intervalle

x \leq 3 or x \geq 5

\frac{2 ^{3} \cdot 3 ^{2}}{( 2 \cdot 3 ) ^{5}}

6 (2 x - 3) = 9 x + 12 + 3 x

x^{2} + 8 x + 11 = 0

∣ 2 x - 8 ∣ = 7

x^{2} + 12 x + 20 = 0