What are the solutions to the equation -9y^2+24=0 ?

The solutions to the equation -9y^2+24=0 are y=(2sqrt(2))/(sqrt(3)),y=-(2sqrt(2))/(sqrt(3))

Find the zeros of -9y^2+24=0

The zeros of -9y^2+24=0 are y=(2sqrt(2))/(sqrt(3)),y=-(2sqrt(2))/(sqrt(3))

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- 9 y^{2} + 24 = 0

y = \frac{2 2}{3}, y = - \frac{2 2}{3}

Decimal Notation

y = 1.63299 \dots, y = - 1.63299 \dots

Solution steps

- 9 y^{2} + 24 = 0

Move

24

to the right side

- 9 y^{2} = - 24

Divide both sides by

- 9

y^{2} = \frac{8}{3}

For

x^{2} = f (a)

the solutions are

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

y = \frac{8}{3}, y = - \frac{8}{3}

\frac{8}{3} = \frac{2 2}{3}

- \frac{8}{3} = - \frac{2 2}{3}

y = \frac{2 2}{3}, y = - \frac{2 2}{3}

x^{2} = 100

x^{2} + x + 24 = 0

x^{2} - 800 = 0

x, 17 x^{2} = 3 x

x, x^{2} = y^{2} - 4

What are the solutions to the equation -9y^2+24=0 ?
The solutions to the equation -9y^2+24=0 are y=(2sqrt(2))/(sqrt(3)),y=-(2sqrt(2))/(sqrt(3))
Find the zeros of -9y^2+24=0
The zeros of -9y^2+24=0 are y=(2sqrt(2))/(sqrt(3)),y=-(2sqrt(2))/(sqrt(3))