What is -x^2+7x-10>= 0 ?

The solution to -x^2+7x-10>= 0 is 2<= x<= 5

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-x^2+7x-10>= 0

- x^{2} + 7 x - 10 \geq 0

2 \leq x \leq 5

Interval Notation

[2, 5]

Solution steps

- x^{2} + 7 x - 10 \geq 0

Factor

- x^{2} + 7 x - 10 : - (x - 2) (x - 5)

- (x - 2) (x - 5) \geq 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 2) (x - 5)) (- 1) \leq 0 \cdot (- 1)

Simplify

(x - 2) (x - 5) \leq 0

Identify the intervals

2 \leq x \leq 5

\frac{( 2 t + 5 ) ( 3 t - 1 )}{3} + \frac{t ^{2} + 5}{2} = \frac{7 t - 5}{6} + 1

∣ 2 x ∣ = x^{2} - 3

4 x^{3} + 12 x^{2} - x - 3

- 1 \cdot \frac{1}{3}

\frac{2}{5} \cdot 6