What is -16t^2+32t+8>= 0 ?

The solution to -16t^2+32t+8>= 0 is -sqrt(3/2)+1<= t<= sqrt(3/2)+1

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-16t^2+32t+8>= 0

- 16 t^{2} + 32 t + 8 \geq 0

- \frac{3}{2} + 1 \leq t \leq \frac{3}{2} + 1

Interval Notation

[- \frac{3}{2} + 1, \frac{3}{2} + 1]

Decimal Notation

- 0.22474 \dots \leq t \leq 2.22474 \dots

Solution steps

- 16 t^{2} + 32 t + 8 \geq 0

Rewrite in standard form

- 2 t^{2} + 4 t + 1 \geq 0

Complete the square

- 2 t^{2} + 4 t + 1 : - 2 (t - 1)^{2} + 3

- 2 (t - 1)^{2} + 3 \geq 0

Move

3

to the right side

- 2 (t - 1)^{2} \geq - 3

Multiply both sides by

- 1

2 (t - 1)^{2} \leq 3

Divide both sides by

2

(t - 1)^{2} \leq \frac{3}{2}

For

u^{n} \leq a

, if

n

is even then

- \frac{3}{2} \leq t - 1 \leq \frac{3}{2}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{3}{2} \leq t - 1 and t - 1 \leq \frac{3}{2}

- \frac{3}{2} \leq t - 1 : t \geq - \frac{3}{2} + 1

t - 1 \leq \frac{3}{2} : t \leq \frac{3}{2} + 1

Combine the intervals

t \geq - \frac{3}{2} + 1 and t \leq \frac{3}{2} + 1

Merge Overlapping Intervals

- \frac{3}{2} + 1 \leq t \leq \frac{3}{2} + 1

∣ 2 t + 1 ∣ + 4 = 5

(3 x^{3} y^{5})^{4}

- 12 + 33

4 (x + 2) = 48

1 - 4 y^{2} = 0

What is -16t^2+32t+8>= 0 ?
The solution to -16t^2+32t+8>= 0 is -sqrt(3/2)+1<= t<= sqrt(3/2)+1