What is 5x+4<= 3x^2 ?

The solution to 5x+4 = (sqrt(73)+5)/6

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5x+4<= 3x^2

5 x + 4 \leq 3 x^{2}

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

Interval Notation

(- \infty, \frac{- 73 + 5}{6}] \cup [\frac{73 + 5}{6}, \infty)

Decimal Notation

x \leq - 0.59066 \dots or x \geq 2.25733 \dots

Solution steps

5 x + 4 \leq 3 x^{2}

Rewrite in standard form

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

Complete the square

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

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

Move

\frac{73}{12}

to the right side

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

Multiply both sides by

- 1

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}

8 + ∣ 4 x + 1 ∣ = 17

10 x^{2} + 23 x - 42 = 0

(\frac{2}{3} \cdot \frac{4}{5}) + \frac{2}{5} - \frac{3}{10}

25 m^{2} - 81 n^{2}

5 x^{2} + 7 x - 2 = 0

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