What is -2x^2+6x<=-15 ?

The solution to -2x^2+6x = (sqrt(39)+3)/2

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-2x^2+6x<=-15

- 2 x^{2} + 6 x \leq - 15

x \leq \frac{- 39 + 3}{2} or x \geq \frac{39 + 3}{2}

Solution steps

- 2 x^{2} + 6 x \leq - 15

Rewrite in standard form

- 2 x^{2} + 6 x + 15 \leq 0

Complete the square

- 2 x^{2} + 6 x + 15 : - 2 (x - \frac{3}{2})^{2} + \frac{39}{2}

- 2 (x - \frac{3}{2})^{2} + \frac{39}{2} \leq 0

Move

\frac{39}{2}

to the right side

- 2 (x - \frac{3}{2})^{2} \leq - \frac{39}{2}

Multiply both sides by

- 1

2 (x - \frac{3}{2})^{2} \geq \frac{39}{2}

Divide both sides by

2

(x - \frac{3}{2})^{2} \geq \frac{39}{4}

For

u^{n} \geq a

, if

n

is even then

x - \frac{3}{2} \leq - \frac{39}{4} or x - \frac{3}{2} \geq \frac{39}{4}

x - \frac{3}{2} \leq - \frac{39}{4} : x \leq \frac{- 39 + 3}{2}

x - \frac{3}{2} \geq \frac{39}{4} : x \geq \frac{39 + 3}{2}

Combine the intervals

x \leq \frac{- 39 + 3}{2} or x \geq \frac{39 + 3}{2}

\frac{1}{4} + \frac{7}{10}

3 (2 x + 3) (5 x - 4) = 0

27 n = 216

2 - \frac{1}{2}

\frac{75}{0.15} = \frac{0.15 \cdot x}{0.15}

What is -2x^2+6x<=-15 ?
The solution to -2x^2+6x<=-15 is x<= (-sqrt(39)+3)/2 \lor x>= (sqrt(39)+3)/2