What is 6n^2+20n<= 50n^3 ?

The solution to 6n^2+20n = (3+sqrt(1009))/(50)

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6n^2+20n<= 50n^3

6 n^{2} + 20 n \leq 50 n^{3}

\frac{3 - 1009}{50} \leq n \leq 0 or n \geq \frac{3 + 1009}{50}

Interval Notation

[\frac{3 - 1009}{50}, 0] \cup [\frac{3 + 1009}{50}, \infty)

Decimal Notation

- 0.57529 \dots \leq n \leq 0 or n \geq 0.69529 \dots

Solution steps

6 n^{2} + 20 n \leq 50 n^{3}

Rewrite in standard form

3 n^{2} + 10 n - 25 n^{3} \leq 0

Factor

3 n^{2} + 10 n - 25 n^{3} : - n (25 n^{2} - 3 n - 10)

- n (25 n^{2} - 3 n - 10) \leq 0

Multiply both sides by

- 1

(reverse the inequality)

(- n (25 n^{2} - 3 n - 10)) (- 1) \geq 0 \cdot (- 1)

Simplify

n (25 n^{2} - 3 n - 10) \geq 0

Identify the intervals

\frac{3 - 1009}{50} \leq n \leq 0 or n \geq \frac{3 + 1009}{50}

- 4 \leq - 2 (x + 8) < 8

3 x^{2} - 6 x + 11 = 0

\frac{ln ( 2 )}{0.12}

∣ x - 2 ∣ + ∣ x + 4 ∣ = 8

3 (x + 1) + 4 (x + 3) = 22

What is 6n^2+20n<= 50n^3 ?
The solution to 6n^2+20n<= 50n^3 is (3-sqrt(1009))/(50)<= n<= 0\lor n>= (3+sqrt(1009))/(50)