What are the solutions to the equation -16x^2+1362=0 ?

The solutions to the equation -16x^2+1362=0 are x=(sqrt(681))/(2sqrt(2)),x=-(sqrt(681))/(2sqrt(2))

Find the zeros of -16x^2+1362=0

The zeros of -16x^2+1362=0 are x=(sqrt(681))/(2sqrt(2)),x=-(sqrt(681))/(2sqrt(2))

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- 16 x^{2} + 1362 = 0

x = \frac{681}{2 2}, x = - \frac{681}{2 2}

Decimal Notation

x = 9.22632 \dots, x = - 9.22632 \dots

Solution steps

- 16 x^{2} + 1362 = 0

Move

1362

to the right side

- 16 x^{2} = - 1362

Divide both sides by

- 16

x^{2} = \frac{681}{8}

For

x^{2} = f (a)

the solutions are

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

x = \frac{681}{8}, x = - \frac{681}{8}

\frac{681}{8} = \frac{681}{2 2}

- \frac{681}{8} = - \frac{681}{2 2}

x = \frac{681}{2 2}, x = - \frac{681}{2 2}

x^{2} + 15 x + 62 = 6

x^{2} - 9 (x - 1)^{2} = 0

x^{2} - 2 x + 7 = 4

1 - 0.5 x + 0.3 x^{2} = 0

6 x^{2} - 18 x - 24 = 62

What are the solutions to the equation -16x^2+1362=0 ?
The solutions to the equation -16x^2+1362=0 are x=(sqrt(681))/(2sqrt(2)),x=-(sqrt(681))/(2sqrt(2))
Find the zeros of -16x^2+1362=0
The zeros of -16x^2+1362=0 are x=(sqrt(681))/(2sqrt(2)),x=-(sqrt(681))/(2sqrt(2))