What is -x^2+x+6>= 0 ?

The solution to -x^2+x+6>= 0 is -2<= x<= 3

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-x^2+x+6>= 0

- x^{2} + x + 6 \geq 0

- 2 \leq x \leq 3

Interval Notation

[- 2, 3]

Solution steps

- x^{2} + x + 6 \geq 0

Factor

- x^{2} + x + 6 : - (x + 2) (x - 3)

- (x + 2) (x - 3) \geq 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x + 2) (x - 3)) (- 1) \leq 0 \cdot (- 1)

Simplify

(x + 2) (x - 3) \leq 0

Identify the intervals

- 2 \leq x \leq 3

lo g_{8} (4) + lo g_{8} (16)

- 3 \leq \frac{3 x + 17}{2} < 2

w^{2} - 144 = 0

x^{2} + 2 x - 99 = 0

4 x^{3} + 16 x^{2} - 25 x - 100