The solution to 2x^2<3x+7 is (-sqrt(65)+3)/4 <x<(sqrt(65)+3)/4

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2x^2<3x+7

Solution

2 x^{2} < 3 x + 7

\frac{- 65 + 3}{4} < x < \frac{65 + 3}{4}

Interval Notation

(\frac{- 65 + 3}{4}, \frac{65 + 3}{4})

Decimal Notation

- 1.26556 \dots < x < 2.76556 \dots

Solution steps

2 x^{2} < 3 x + 7

Rewrite in standard form

2 x^{2} - 7 - 3 x < 0

Complete the square

2 x^{2} - 7 - 3 x : 2 (x - \frac{3}{4})^{2} - \frac{65}{8}

2 (x - \frac{3}{4})^{2} - \frac{65}{8} < 0

Move

\frac{65}{8}

to the right side

2 (x - \frac{3}{4})^{2} < \frac{65}{8}

Divide both sides by

2

(x - \frac{3}{4})^{2} < \frac{65}{16}

For

u^{n} < a

, if

n

is even then

- \frac{65}{16} < x - \frac{3}{4} < \frac{65}{16}

a < u < b

then

a < u and u < b

- \frac{65}{16} < x - \frac{3}{4} and x - \frac{3}{4} < \frac{65}{16}

- \frac{65}{16} < x - \frac{3}{4} : x > \frac{- 65 + 3}{4}

x - \frac{3}{4} < \frac{65}{16} : x < \frac{65 + 3}{4}

Combine the intervals

x > \frac{- 65 + 3}{4} and x < \frac{65 + 3}{4}

Merge Overlapping Intervals

\frac{- 65 + 3}{4} < x < \frac{65 + 3}{4}

0 = - 5 x - 5

j^{2} + 7 j + 10

x + 5 + (x + 5)^{2}

3 u^{2} + 8 u + 5

∣ 2 x - 3 ∣ = ∣ 4 x + 2 ∣

What is 2x^2<3x+7 ?
The solution to 2x^2<3x+7 is (-sqrt(65)+3)/4 <x<(sqrt(65)+3)/4