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

The solution to -x^2-x+2>= 0 is -2<= x<= 1

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-x^2-x+2>= 0

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

- 2 \leq x \leq 1

Interval Notation

[- 2, 1]

Solution steps

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

Factor

- x^{2} - x + 2 : - (x - 1) (x + 2)

- (x - 1) (x + 2) \geq 0

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

Identify the intervals

- 2 \leq x \leq 1

e^{2} = x

5 x^{2} + 31 x - 28

x^{2} = 13 + 6 x

{6 + [9 - (- 5 + 10) + (7 - 3)] + [(- 3) - (- 12) + 13]}

\frac{1}{2 + \frac{1}{1 + \frac{1}{2}}}