What is (3x-4)(2x+2)<= 0 ?

The solution to (3x-4)(2x+2)<= 0 is -1<= x<= 4/3

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(3x-4)(2x+2)<= 0

(3 x - 4) (2 x + 2) \leq 0

- 1 \leq x \leq \frac{4}{3}

Interval Notation

[- 1, \frac{4}{3}]

Decimal Notation

- 1 \leq x \leq 1.33333 \dots

Solution steps

(3 x - 4) (2 x + 2) \leq 0

Rewrite in standard form

3 x^{2} - x - 4 \leq 0

Factor

3 x^{2} - x - 4 : (x + 1) (3 x - 4)

(x + 1) (3 x - 4) \leq 0

Identify the intervals

- 1 \leq x \leq \frac{4}{3}

4 x - 13 x = 18

x^{2} - 4 x - 8 = 0

x^{2} - 10 x + 25 - y^{2}

\frac{1}{3} (x + 4) - \frac{5}{6} (x - 3) > \frac{1}{2} x + 1

x^{2} - 12 \leq 4 x