What are the solutions to the equation 16(x^2-2)+13=-10 ?

The solutions to the equation 16(x^2-2)+13=-10 are x= 3/4 ,x=-3/4

Find the zeros of 16(x^2-2)+13=-10

The zeros of 16(x^2-2)+13=-10 are x= 3/4 ,x=-3/4

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16 (x^{2} - 2) + 13 = - 10

x = \frac{3}{4}, x = - \frac{3}{4}

Decimal Notation

x = 0.75, x = - 0.75

Solution steps

16 (x^{2} - 2) + 13 = - 10

Move

13

to the right side

16 (x^{2} - 2) = - 23

Divide both sides by

16

x^{2} - 2 = - \frac{23}{16}

Move

2

to the right side

x^{2} = \frac{9}{16}

For

x^{2} = f (a)

the solutions are

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

x = \frac{9}{16}, x = - \frac{9}{16}

\frac{9}{16} = \frac{3}{4}

- \frac{9}{16} = - \frac{3}{4}

x = \frac{3}{4}, x = - \frac{3}{4}

4 x^{2} - 9 x + 1 = 0

9 x^{2} - 9 x - 4 = 0

3 x^{2} + 12 x - 36 = 0

λ^{2} - 4 λ - 5 = 0

x^{2} - 22 x + 112 = 0

What are the solutions to the equation 16(x^2-2)+13=-10 ?
The solutions to the equation 16(x^2-2)+13=-10 are x= 3/4 ,x=-3/4
Find the zeros of 16(x^2-2)+13=-10
The zeros of 16(x^2-2)+13=-10 are x= 3/4 ,x=-3/4