What is 2x^2-x-6<14 ?

The solution to 2x^2-x-6<14 is (-sqrt(161)+1)/4 <x<(sqrt(161)+1)/4

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2x^2-x-6<14

2 x^{2} - x - 6 < 14

\frac{- 161 + 1}{4} < x < \frac{161 + 1}{4}

Interval Notation

(\frac{- 161 + 1}{4}, \frac{161 + 1}{4})

Decimal Notation

- 2.92214 \dots < x < 3.42214 \dots

Solution steps

2 x^{2} - x - 6 < 14

Rewrite in standard form

2 x^{2} - x - 20 < 0

Complete the square

2 x^{2} - x - 20 : 2 (x - \frac{1}{4})^{2} - \frac{161}{8}

2 (x - \frac{1}{4})^{2} - \frac{161}{8} < 0

Move

\frac{161}{8}

to the right side

2 (x - \frac{1}{4})^{2} < \frac{161}{8}

Divide both sides by

2

(x - \frac{1}{4})^{2} < \frac{161}{16}

For

u^{n} < a

, if

n

is even then

- \frac{161}{16} < x - \frac{1}{4} < \frac{161}{16}

a < u < b

then

a < u and u < b

- \frac{161}{16} < x - \frac{1}{4} and x - \frac{1}{4} < \frac{161}{16}

- \frac{161}{16} < x - \frac{1}{4} : x > \frac{- 161 + 1}{4}

x - \frac{1}{4} < \frac{161}{16} : x < \frac{161 + 1}{4}

Combine the intervals

x > \frac{- 161 + 1}{4} and x < \frac{161 + 1}{4}

Merge Overlapping Intervals

\frac{- 161 + 1}{4} < x < \frac{161 + 1}{4}

\frac{1}{( x ^{2} )} < 2

x (x - 3) (x - 1)^{2} < 0

\frac{( 3 x - 2 )}{( x - 1 )} \geq 0

(x - 3) (x + 1) (x + 2) (x - 9) \geq 0

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

What is 2x^2-x-6<14 ?
The solution to 2x^2-x-6<14 is (-sqrt(161)+1)/4 <x<(sqrt(161)+1)/4