What are the solutions to the equation -16t^2+96=0 ?

The solutions to the equation -16t^2+96=0 are t=sqrt(6),t=-sqrt(6)

Find the zeros of -16t^2+96=0

The zeros of -16t^2+96=0 are t=sqrt(6),t=-sqrt(6)

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- 16 t^{2} + 96 = 0

t = 6, t = - 6

Decimal Notation

t = 2.44948 \dots, t = - 2.44948 \dots

Solution steps

- 16 t^{2} + 96 = 0

Move

96

to the right side

- 16 t^{2} = - 96

Divide both sides by

- 16

t^{2} = 6

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

t = 6, t = - 6

- 2 = - \frac{1}{4} x - \frac{1}{2}

\frac{a}{12 b} + \frac{b}{8 a}

2 (m - 3) + 7 < 21

(s^{\frac{3}{2}})^{\frac{4}{7}}

- 2 < x \leq 7

What are the solutions to the equation -16t^2+96=0 ?
The solutions to the equation -16t^2+96=0 are t=sqrt(6),t=-sqrt(6)
Find the zeros of -16t^2+96=0
The zeros of -16t^2+96=0 are t=sqrt(6),t=-sqrt(6)