What is (x+2)(2x-1)(3x+7)>= 0 ?

The solution to (x+2)(2x-1)(3x+7)>= 0 is -7/3 = 1/2

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(x+2)(2x-1)(3x+7)>= 0

(x + 2) (2 x - 1) (3 x + 7) \geq 0

- \frac{7}{3} \leq x \leq - 2 or x \geq \frac{1}{2}

Interval Notation

[- \frac{7}{3}, - 2] \cup [\frac{1}{2}, \infty)

Decimal Notation

- 2.33333 \dots \leq x \leq - 2 or x \geq 0.5

Solution steps

(x + 2) (2 x - 1) (3 x + 7) \geq 0

Identify the intervals

- \frac{7}{3} \leq x \leq - 2 or x \geq \frac{1}{2}

\frac{( x + 2 ) ( 2 x - 3 )}{( x - 3 ) ( x + 1 )} \leq 6

(p - 3) (p - 1) = 0

3 t^{2} - 3 t + 1 = 0

- 9.79 + 7.8 v = - 17.39 - 11.1 v + 18.4 v

\frac{5 - \frac{6}{4} x}{\frac{1}{2} x - 3} - \frac{5}{6} \leq 6 + 2 x^{0}

What is (x+2)(2x-1)(3x+7)>= 0 ?
The solution to (x+2)(2x-1)(3x+7)>= 0 is -7/3 <= x<=-2\lor x>= 1/2