What is 3x^2-7x-4>= 0 ?

The solution to 3x^2-7x-4>= 0 is x = (sqrt(97)+7)/6

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3x^2-7x-4>= 0

3 x^{2} - 7 x - 4 \geq 0

x \leq \frac{- 97 + 7}{6} or x \geq \frac{97 + 7}{6}

Interval Notation

(- \infty, \frac{- 97 + 7}{6}] \cup [\frac{97 + 7}{6}, \infty)

Decimal Notation

x \leq - 0.47480 \dots or x \geq 2.80814 \dots

Solution steps

3 x^{2} - 7 x - 4 \geq 0

Complete the square

3 x^{2} - 7 x - 4 : 3 (x - \frac{7}{6})^{2} - \frac{97}{12}

3 (x - \frac{7}{6})^{2} - \frac{97}{12} \geq 0

Move

\frac{97}{12}

to the right side

3 (x - \frac{7}{6})^{2} \geq \frac{97}{12}

Divide both sides by

3

(x - \frac{7}{6})^{2} \geq \frac{97}{36}

For

u^{n} \geq a

, if

n

is even then

x - \frac{7}{6} \leq - \frac{97}{36} or x - \frac{7}{6} \geq \frac{97}{36}

x - \frac{7}{6} \leq - \frac{97}{36} : x \leq \frac{- 97 + 7}{6}

x - \frac{7}{6} \geq \frac{97}{36} : x \geq \frac{97 + 7}{6}

Combine the intervals

x \leq \frac{- 97 + 7}{6} or x \geq \frac{97 + 7}{6}

∣ 2 x + 1 ∣ \geq - 9

- 3 u + 7 = - 9 u + 49

(85)^{2}

- \frac{t}{4} = - 3 π

3 n + 46 = 1 + 8 n

What is 3x^2-7x-4>= 0 ?
The solution to 3x^2-7x-4>= 0 is x<= (-sqrt(97)+7)/6 \lor x>= (sqrt(97)+7)/6