What are the solutions to the equation -16t^2+25=0 ?

The solutions to the equation -16t^2+25=0 are t= 5/4 ,t=-5/4

Find the zeros of -16t^2+25=0

The zeros of -16t^2+25=0 are t= 5/4 ,t=-5/4

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- 16 t^{2} + 25 = 0

t = \frac{5}{4}, t = - \frac{5}{4}

Decimal Notation

t = 1.25, t = - 1.25

Solution steps

- 16 t^{2} + 25 = 0

Move

25

to the right side

- 16 t^{2} = - 25

Divide both sides by

- 16

t^{2} = \frac{25}{16}

For

x^{2} = f (a)

the solutions are

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

t = \frac{25}{16}, t = - \frac{25}{16}

\frac{25}{16} = \frac{5}{4}

- \frac{25}{16} = - \frac{5}{4}

t = \frac{5}{4}, t = - \frac{5}{4}

3 x^{2} + 6 x + 1 = 0

- 2 x^{2} - 9 x - 5 = 0

x, h^{2} = x^{2} + 2 y^{2}

4 x^{2} + a^{2} = 4 a x + b^{2}

x^{2} - 10 x - 66 = 0

What are the solutions to the equation -16t^2+25=0 ?
The solutions to the equation -16t^2+25=0 are t= 5/4 ,t=-5/4
Find the zeros of -16t^2+25=0
The zeros of -16t^2+25=0 are t= 5/4 ,t=-5/4