What is 3x^2+5x-1<= 5x^2-6 ?

The solution to 3x^2+5x-1 = (sqrt(65)+5)/4

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3x^2+5x-1<= 5x^2-6

3 x^{2} + 5 x - 1 \leq 5 x^{2} - 6

x \leq \frac{- 65 + 5}{4} or x \geq \frac{65 + 5}{4}

Interval Notation

(- \infty, \frac{- 65 + 5}{4}] \cup [\frac{65 + 5}{4}, \infty)

Decimal Notation

x \leq - 0.76556 \dots or x \geq 3.26556 \dots

Solution steps

3 x^{2} + 5 x - 1 \leq 5 x^{2} - 6

Rewrite in standard form

- 2 x^{2} + 5 x + 5 \leq 0

Complete the square

- 2 x^{2} + 5 x + 5 : - 2 (x - \frac{5}{4})^{2} + \frac{65}{8}

- 2 (x - \frac{5}{4})^{2} + \frac{65}{8} \leq 0

Move

\frac{65}{8}

to the right side

- 2 (x - \frac{5}{4})^{2} \leq - \frac{65}{8}

Multiply both sides by

- 1

2 (x - \frac{5}{4})^{2} \geq \frac{65}{8}

Divide both sides by

2

(x - \frac{5}{4})^{2} \geq \frac{65}{16}

For

u^{n} \geq a

, if

n

is even then

x - \frac{5}{4} \leq - \frac{65}{16} or x - \frac{5}{4} \geq \frac{65}{16}

x - \frac{5}{4} \leq - \frac{65}{16} : x \leq \frac{- 65 + 5}{4}

x - \frac{5}{4} \geq \frac{65}{16} : x \geq \frac{65 + 5}{4}

Combine the intervals

x \leq \frac{- 65 + 5}{4} or x \geq \frac{65 + 5}{4}

- 2.5 x^{2} + 400 x + 3600 = 0

20 x^{2} - 17 x - 63

3 i + 2 i

x^{2} > x + 2

\frac{( n + 1 ) !}{n !}

What is 3x^2+5x-1<= 5x^2-6 ?
The solution to 3x^2+5x-1<= 5x^2-6 is x<= (-sqrt(65)+5)/4 \lor x>= (sqrt(65)+5)/4