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

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

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2n^2+5n-1<0

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

\frac{- 5 - 33}{4} < n < \frac{- 5 + 33}{4}

Interval Notation

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

Decimal Notation

- 2.68614 \dots < n < 0.18614 \dots

Solution steps

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

Identify the intervals

\frac{- 5 - 33}{4} < n < \frac{- 5 + 33}{4}

(x + 4)^{2} = x^{2} + 8

2 w^{2} + 9 w + 9

(\frac{2 c - 3}{5})^{2} + 2 (\frac{2 c - 3}{5}) = 8

e^{l n (12)}

16 a^{2} - 121

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