What is -x^2+10x-21-(3x-15)<= 0 ?

The solution to -x^2+10x-21-(3x-15) = 6

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-x^2+10x-21-(3x-15)<= 0

- x^{2} + 10 x - 21 - (3 x - 15) \leq 0

x \leq 1 or x \geq 6

Interval Notation

(- \infty, 1] \cup [6, \infty)

Solution steps

- x^{2} + 10 x - 21 - (3 x - 15) \leq 0

Factor

- x^{2} + 10 x - 21 - (3 x - 15) : - (x - 1) (x - 6)

- (x - 1) (x - 6) \leq 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 1) (x - 6)) (- 1) \geq 0 \cdot (- 1)

Simplify

(x - 1) (x - 6) \geq 0

Identify the intervals

x \leq 1 or x \geq 6

ln ∣ 2 ∣

3 x^{4} - 432 x^{2}

12 x^{2} - 8 x

2 j < 10

12 x^{2} - 2 x - 4

What is -x^2+10x-21-(3x-15)<= 0 ?
The solution to -x^2+10x-21-(3x-15)<= 0 is x<= 1\lor x>= 6