What is -x^2-8x-12>= 0 ?

The solution to -x^2-8x-12>= 0 is -6<= x<=-2

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-x^2-8x-12>= 0

- x^{2} - 8 x - 12 \geq 0

- 6 \leq x \leq - 2

Interval Notation

[- 6, - 2]

Solution steps

- x^{2} - 8 x - 12 \geq 0

Factor

- x^{2} - 8 x - 12 : - (x + 2) (x + 6)

- (x + 2) (x + 6) \geq 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x + 2) (x + 6)) (- 1) \leq 0 \cdot (- 1)

Simplify

(x + 2) (x + 6) \leq 0

Identify the intervals

- 6 \leq x \leq - 2

18 0^{2}

p (x) \frac{x + 2}{x ^{2} + x}

(\frac{1}{2})^{- \frac{1}{2}}

2 x^{2} + 11 x + 5

x^{3} + 14 x^{2} + 49 x