What are the solutions to the equation (9z+1/2)^2+61=39 ?

The solutions to the equation (9z+1/2)^2+61=39 are z=-1/18+i(sqrt(22))/9 ,z=-1/18-i(sqrt(22))/9

Find the zeros of (9z+1/2)^2+61=39

The zeros of (9z+1/2)^2+61=39 are z=-1/18+i(sqrt(22))/9 ,z=-1/18-i(sqrt(22))/9

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(9 z + \frac{1}{2})^{2} + 61 = 39

z = - \frac{1}{18} + i \frac{22}{9}, z = - \frac{1}{18} - i \frac{22}{9}

Solution steps

(9 z + \frac{1}{2})^{2} + 61 = 39

Move

61

to the right side

(9 z + \frac{1}{2})^{2} = - 22

For

(g (x))^{2} = f (a)

the solutions are

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

Solve

9 z + \frac{1}{2} = - 22 : z = - \frac{1}{18} + i \frac{22}{9}

Solve

9 z + \frac{1}{2} = - - 22 : z = - \frac{1}{18} - i \frac{22}{9}

The solutions to the quadratic equation are:

z = - \frac{1}{18} + i \frac{22}{9}, z = - \frac{1}{18} - i \frac{22}{9}

- 40 m^{2} + 10 m - 1 = 0

m^{2} - 11 m + 30 = 0

x^{2} - 5 x - 24 = 0

x^{2} + 8 x + 60 = 0

x, x^{2} - 5 x - 14 = 0

What are the solutions to the equation (9z+1/2)^2+61=39 ?
The solutions to the equation (9z+1/2)^2+61=39 are z=-1/18+i(sqrt(22))/9 ,z=-1/18-i(sqrt(22))/9
Find the zeros of (9z+1/2)^2+61=39
The zeros of (9z+1/2)^2+61=39 are z=-1/18+i(sqrt(22))/9 ,z=-1/18-i(sqrt(22))/9