What is -12x^2+3x+1<0 ?

The solution to -12x^2+3x+1 (sqrt(19))/(8sqrt(3))+1/8

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-12x^2+3x+1<0

- 12 x^{2} + 3 x + 1 < 0

x < - \frac{19}{8 3} + \frac{1}{8} or x > \frac{19}{8 3} + \frac{1}{8}

Interval Notation

(- \infty, - \frac{19}{8 3} + \frac{1}{8}) \cup (\frac{19}{8 3} + \frac{1}{8}, \infty)

Decimal Notation

x < - 0.18957 \dots or x > 0.43957 \dots

Solution steps

- 12 x^{2} + 3 x + 1 < 0

Complete the square

- 12 x^{2} + 3 x + 1 : - 12 (x - \frac{1}{8})^{2} + \frac{19}{16}

- 12 (x - \frac{1}{8})^{2} + \frac{19}{16} < 0

Move

\frac{19}{16}

to the right side

- 12 (x - \frac{1}{8})^{2} < - \frac{19}{16}

Multiply both sides by

- 1

12 (x - \frac{1}{8})^{2} > \frac{19}{16}

Divide both sides by

12

(x - \frac{1}{8})^{2} > \frac{19}{192}

For

u^{n} > a

, if

n

is even then

x - \frac{1}{8} < - \frac{19}{192} or x - \frac{1}{8} > \frac{19}{192}

x - \frac{1}{8} < - \frac{19}{192} : x < - \frac{19}{8 3} + \frac{1}{8}

x - \frac{1}{8} > \frac{19}{192} : x > \frac{19}{8 3} + \frac{1}{8}

Combine the intervals

x < - \frac{19}{8 3} + \frac{1}{8} or x > \frac{19}{8 3} + \frac{1}{8}

\frac{ln ( 4 )}{2}

\frac{1}{6} (2 x^{4} + 3 x^{3} - 16 x - 24)^{2}

2 y + 3 > 9

(4 x - 3)^{2} = 14

∣ x - 5 ∣ + 1 \leq 2 x

What is -12x^2+3x+1<0 ?
The solution to -12x^2+3x+1<0 is x<-(sqrt(19))/(8sqrt(3))+1/8 \lor x>(sqrt(19))/(8sqrt(3))+1/8