What is -x^2+10x-21<= 0 ?

The solution to -x^2+10x-21 = 7

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-x^2+10x-21<= 0

- x^{2} + 10 x - 21 \leq 0

x \leq 3 or x \geq 7

Interval Notation

(- \infty, 3] \cup [7, \infty)

Solution steps

- x^{2} + 10 x - 21 \leq 0

Factor

- x^{2} + 10 x - 21 : - (x - 3) (x - 7)

- (x - 3) (x - 7) \leq 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 3) (x - 7)) (- 1) \geq 0 \cdot (- 1)

Simplify

(x - 3) (x - 7) \geq 0

Identify the intervals

x \leq 3 or x \geq 7

\frac{1}{( ( n - 2 ) ( n - 3 ) )} < 0

x^{2} + 2 x < 35

4 - (\frac{3}{x}) < x

\frac{( 2 x ^{2} - 1 )}{( 2 x ^{2} + 7 x - 15 )} > 1

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