What is 3x^2-5x-4>= 0 ?

The solution to 3x^2-5x-4>= 0 is x = (sqrt(73)+5)/6

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3x^2-5x-4>= 0

3 x^{2} - 5 x - 4 \geq 0

x \leq \frac{- 73 + 5}{6} or x \geq \frac{73 + 5}{6}

Solution steps

3 x^{2} - 5 x - 4 \geq 0

Complete the square

3 x^{2} - 5 x - 4 : 3 (x - \frac{5}{6})^{2} - \frac{73}{12}

3 (x - \frac{5}{6})^{2} - \frac{73}{12} \geq 0

Move

\frac{73}{12}

to the right side

3 (x - \frac{5}{6})^{2} \geq \frac{73}{12}

Divide both sides by

3

(x - \frac{5}{6})^{2} \geq \frac{73}{36}

For

u^{n} \geq a

, if

n

is even then

x - \frac{5}{6} \leq - \frac{73}{36} or x - \frac{5}{6} \geq \frac{73}{36}

x - \frac{5}{6} \leq - \frac{73}{36} : x \leq \frac{- 73 + 5}{6}

x - \frac{5}{6} \geq \frac{73}{36} : x \geq \frac{73 + 5}{6}

Combine the intervals

x \leq \frac{- 73 + 5}{6} or x \geq \frac{73 + 5}{6}

(x + 6)^{2} (x + 2) (x - 1) > 0

5 x - 3 \leq 12

- 5 k^{2} - 8 k + 3 = 0

\frac{4}{5} (\frac{3}{4} - \frac{3}{6})

70 - a \geq 25

What is 3x^2-5x-4>= 0 ?
The solution to 3x^2-5x-4>= 0 is x<= (-sqrt(73)+5)/6 \lor x>= (sqrt(73)+5)/6