What is 7x+5x<x^2-10 ?

The solution to 7x+5x sqrt(46)+6

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7x+5x<x^2-10

7 x + 5 x < x^{2} - 10

x < - 46 + 6 or x > 46 + 6

Interval Notation

(- \infty, - 46 + 6) \cup (46 + 6, \infty)

Decimal Notation

x < - 0.78232 \dots or x > 12.78232 \dots

Solution steps

7 x + 5 x < x^{2} - 10

Rewrite in standard form

12 x + 10 - x^{2} < 0

Complete the square

12 x + 10 - x^{2} : - (x - 6)^{2} + 46

- (x - 6)^{2} + 46 < 0

Move

46

to the right side

- (x - 6)^{2} < - 46

Multiply both sides by

- 1

(x - 6)^{2} > 46

For

u^{n} > a

, if

n

is even then

x - 6 < - 46 or x - 6 > 46

x - 6 < - 46 : x < - 46 + 6

x - 6 > 46 : x > 46 + 6

Combine the intervals

x < - 46 + 6 or x > 46 + 6

\frac{( x - 2 ) ^{2} ( x - 1 )}{( x - 3 )} \geq 0

\frac{1}{( n - 2 )} + \frac{1}{( n - 1 )} > \frac{1}{n}

(x - 2) (x - 3) (x + 2) (x + 3) > 0

x (9 x - 4) < (3 x + 2)^{2}

(x - 1)^{2021} (x + 1)^{2022} < 0

What is 7x+5x<x^2-10 ?
The solution to 7x+5x<x^2-10 is x<-sqrt(46)+6\lor x>sqrt(46)+6