What is 128+16t-16t^2>= 32 ?

The solution to 128+16t-16t^2>= 32 is -2<= t<= 3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

128+16t-16t^2>= 32

128 + 16 t - 16 t^{2} \geq 32

- 2 \leq t \leq 3

Interval Notation

[- 2, 3]

Solution steps

128 + 16 t - 16 t^{2} \geq 32

Rewrite in standard form

- t^{2} + t + 6 \geq 0

Factor

- t^{2} + t + 6 : - (t + 2) (t - 3)

- (t + 2) (t - 3) \geq 0

Multiply both sides by

- 1

(reverse the inequality)

(- (t + 2) (t - 3)) (- 1) \leq 0 \cdot (- 1)

Simplify

(t + 2) (t - 3) \leq 0

Identify the intervals

- 2 \leq t \leq 3

8 w + 50 = 142

45 x^{2} y^{5} z^{8}

\frac{n ^{2} - 225}{n ^{2} + 30 n + 225}

(\frac{y ^{2}}{y ^{\frac{4}{5}}})^{\frac{1}{3}}

x^{8} - 16 x^{4}