What are the solutions to the equation 4k^2-7=29 ?

The solutions to the equation 4k^2-7=29 are k=3,k=-3

Find the zeros of 4k^2-7=29

The zeros of 4k^2-7=29 are k=3,k=-3

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4 k^{2} - 7 = 29

k = 3, k = - 3

Solution steps

4 k^{2} - 7 = 29

Move

7

to the right side

4 k^{2} = 36

Divide both sides by

4

k^{2} = 9

For

x^{2} = f (a)

the solutions are

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

k = 9, k = - 9

9 = 3

- 9 = - 3

k = 3, k = - 3

x, x^{2} - 6 x - 9 y^{2} - 54 y - 72 = 0

6 t^{2} + 6 t + 14 = 2

0 = 40 x - 4.9 x^{2}

m^{2} = 5 m - 6

x^{2} + (2 x)^{2} = 1

What are the solutions to the equation 4k^2-7=29 ?
The solutions to the equation 4k^2-7=29 are k=3,k=-3
Find the zeros of 4k^2-7=29
The zeros of 4k^2-7=29 are k=3,k=-3