What is 20-15t^2>= 0 ?

The solution to 20-15t^2>= 0 is -2/(sqrt(3))<= t<= 2/(sqrt(3))

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20-15t^2>= 0

20 - 15 t^{2} \geq 0

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

Interval Notation

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

Decimal Notation

- 1.15470 \dots \leq t \leq 1.15470 \dots

Solution steps

20 - 15 t^{2} \geq 0

Move

20

to the right side

- 15 t^{2} \geq - 20

Multiply both sides by

- 1

15 t^{2} \leq 20

Divide both sides by

15

t^{2} \leq \frac{4}{3}

For

u^{n} \leq a

, if

n

is even then

- \frac{4}{3} \leq t \leq \frac{4}{3}

\frac{4}{3} = \frac{2}{3}

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

20 0^{\frac{1}{2}}

x^{2} - 4 x - 5 \leq 7

8 x^{2} - 12 x + 4 = 0

∣ x + 5 ∣ + 4 < 12

lo g_{5} (- 1)

What is 20-15t^2>= 0 ?
The solution to 20-15t^2>= 0 is -2/(sqrt(3))<= t<= 2/(sqrt(3))