What is 2x^2-3x-15<= 0 ?

The solution to 2x^2-3x-15<= 0 is (-sqrt(129)+3)/4 <= x<= (sqrt(129)+3)/4

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2x^2-3x-15<= 0

2 x^{2} - 3 x - 15 \leq 0

\frac{- 129 + 3}{4} \leq x \leq \frac{129 + 3}{4}

Interval Notation

[\frac{- 129 + 3}{4}, \frac{129 + 3}{4}]

Decimal Notation

- 2.08945 \dots \leq x \leq 3.58945 \dots

Solution steps

2 x^{2} - 3 x - 15 \leq 0

Complete the square

2 x^{2} - 3 x - 15 : 2 (x - \frac{3}{4})^{2} - \frac{129}{8}

2 (x - \frac{3}{4})^{2} - \frac{129}{8} \leq 0

Move

\frac{129}{8}

to the right side

2 (x - \frac{3}{4})^{2} \leq \frac{129}{8}

Divide both sides by

2

(x - \frac{3}{4})^{2} \leq \frac{129}{16}

For

u^{n} \leq a

, if

n

is even then

- \frac{129}{16} \leq x - \frac{3}{4} \leq \frac{129}{16}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{129}{16} \leq x - \frac{3}{4} and x - \frac{3}{4} \leq \frac{129}{16}

- \frac{129}{16} \leq x - \frac{3}{4} : x \geq \frac{- 129 + 3}{4}

x - \frac{3}{4} \leq \frac{129}{16} : x \leq \frac{129 + 3}{4}

Combine the intervals

x \geq \frac{- 129 + 3}{4} and x \leq \frac{129 + 3}{4}

Merge Overlapping Intervals

\frac{- 129 + 3}{4} \leq x \leq \frac{129 + 3}{4}

x^{2} + 2 = 4 - x^{2}

x^{3} - 729 a^{3}

\frac{2}{5} = \frac{1}{3} + f

2 x^{2} + x - 8 = 0

2 x^{2} + 6 x = - 4

What is 2x^2-3x-15<= 0 ?
The solution to 2x^2-3x-15<= 0 is (-sqrt(129)+3)/4 <= x<= (sqrt(129)+3)/4