The solution to 2x^2+3x<4 is (-sqrt(41)-3)/4 <x<(sqrt(41)-3)/4

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2x^2+3x<4

Solution

2 x^{2} + 3 x < 4

\frac{- 41 - 3}{4} < x < \frac{41 - 3}{4}

Interval Notation

(\frac{- 41 - 3}{4}, \frac{41 - 3}{4})

Decimal Notation

- 2.35078 \dots < x < 0.85078 \dots

Solution steps

2 x^{2} + 3 x < 4

Rewrite in standard form

2 x^{2} + 3 x - 4 < 0

Complete the square

2 x^{2} + 3 x - 4 : 2 (x + \frac{3}{4})^{2} - \frac{41}{8}

2 (x + \frac{3}{4})^{2} - \frac{41}{8} < 0

Move

\frac{41}{8}

to the right side

2 (x + \frac{3}{4})^{2} < \frac{41}{8}

Divide both sides by

2

(x + \frac{3}{4})^{2} < \frac{41}{16}

For

u^{n} < a

, if

n

is even then

- \frac{41}{16} < x + \frac{3}{4} < \frac{41}{16}

a < u < b

then

a < u and u < b

- \frac{41}{16} < x + \frac{3}{4} and x + \frac{3}{4} < \frac{41}{16}

- \frac{41}{16} < x + \frac{3}{4} : x > \frac{- 41 - 3}{4}

x + \frac{3}{4} < \frac{41}{16} : x < \frac{41 - 3}{4}

Combine the intervals

x > \frac{- 41 - 3}{4} and x < \frac{41 - 3}{4}

Merge Overlapping Intervals

\frac{- 41 - 3}{4} < x < \frac{41 - 3}{4}

\frac{4}{( x - 2 )} > \frac{3}{( x + 5 )}

2 x^{3} - 3 x^{2} - 8 x - 3 \leq 0

3 x + 4 < x^{2} + 2

2 x^{3} + 4 x^{2} - 24 x + 18 \geq 0

(x - 5) (x - 6) \leq 0

What is 2x^2+3x<4 ?
The solution to 2x^2+3x<4 is (-sqrt(41)-3)/4 <x<(sqrt(41)-3)/4