What is 2-(5-x)2x-6>0 ?

The solution to 2-(5-x)2x-6>0 is x (sqrt(33)+5)/2

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2-(5-x)2x-6>0

2 - (5 - x) 2 x - 6 > 0

x < \frac{- 33 + 5}{2} or x > \frac{33 + 5}{2}

Interval Notation

(- \infty, \frac{- 33 + 5}{2}) \cup (\frac{33 + 5}{2}, \infty)

Decimal Notation

x < - 0.37228 \dots or x > 5.37228 \dots

Solution steps

2 - (5 - x) \cdot 2 x - 6 > 0

Rewrite in standard form

x^{2} - 5 x - 2 > 0

Complete the square

x^{2} - 5 x - 2 : (x - \frac{5}{2})^{2} - \frac{33}{4}

(x - \frac{5}{2})^{2} - \frac{33}{4} > 0

Move

\frac{33}{4}

to the right side

(x - \frac{5}{2})^{2} > \frac{33}{4}

For

u^{n} > a

, if

n

is even then

x - \frac{5}{2} < - \frac{33}{4} or x - \frac{5}{2} > \frac{33}{4}

x - \frac{5}{2} < - \frac{33}{4} : x < \frac{- 33 + 5}{2}

x - \frac{5}{2} > \frac{33}{4} : x > \frac{33 + 5}{2}

Combine the intervals

x < \frac{- 33 + 5}{2} or x > \frac{33 + 5}{2}

5 x^{6} - x^{4} > 5 x^{5}

\frac{1}{x} + \frac{1}{x} - 1 + \frac{1}{x} - 2 > 0

x^{2} + 4 x + 1 \geq - 3

n^{4} + 6 n^{3} + 6 n^{2} + 6 n + 5 < 0

\frac{( 2 x - 3 )}{( x - 1 )} \leq 2

What is 2-(5-x)2x-6>0 ?
The solution to 2-(5-x)2x-6>0 is x<(-sqrt(33)+5)/2 \lor x>(sqrt(33)+5)/2