What is 2x^2+5x-1<0 ?

The solution to 2x^2+5x-1<0 is (-sqrt(33)-5)/4 <x<(sqrt(33)-5)/4

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2x^2+5x-1<0

2 x^{2} + 5 x - 1 < 0

\frac{- 33 - 5}{4} < x < \frac{33 - 5}{4}

Interval Notation

(\frac{- 33 - 5}{4}, \frac{33 - 5}{4})

Decimal Notation

- 2.68614 \dots < x < 0.18614 \dots

Solution steps

2 x^{2} + 5 x - 1 < 0

Complete the square

2 x^{2} + 5 x - 1 : 2 (x + \frac{5}{4})^{2} - \frac{33}{8}

2 (x + \frac{5}{4})^{2} - \frac{33}{8} < 0

Move

\frac{33}{8}

to the right side

2 (x + \frac{5}{4})^{2} < \frac{33}{8}

Divide both sides by

2

(x + \frac{5}{4})^{2} < \frac{33}{16}

For

u^{n} < a

, if

n

is even then

- \frac{33}{16} < x + \frac{5}{4} < \frac{33}{16}

a < u < b

then

a < u and u < b

- \frac{33}{16} < x + \frac{5}{4} and x + \frac{5}{4} < \frac{33}{16}

- \frac{33}{16} < x + \frac{5}{4} : x > \frac{- 33 - 5}{4}

x + \frac{5}{4} < \frac{33}{16} : x < \frac{33 - 5}{4}

Combine the intervals

x > \frac{- 33 - 5}{4} and x < \frac{33 - 5}{4}

Merge Overlapping Intervals

\frac{- 33 - 5}{4} < x < \frac{33 - 5}{4}

59 - 7 q - 3 q^{2} = \frac{1}{3} q^{2} + 8

\frac{y ^{- 3}}{y ^{3}}

\frac{3}{5} x = \frac{7}{10}

4 x^{2} - 21 x y - 18 y^{2}

\frac{7 x ^{2}}{4 y ^{3}} \cdot \frac{2 y}{x}

What is 2x^2+5x-1<0 ?
The solution to 2x^2+5x-1<0 is (-sqrt(33)-5)/4 <x<(sqrt(33)-5)/4