What are the solutions to the equation (3z-18)^2+58=40 ?

The solutions to the equation (3z-18)^2+58=40 are z=sqrt(2)i+6,z=-sqrt(2)i+6

Find the zeros of (3z-18)^2+58=40

The zeros of (3z-18)^2+58=40 are z=sqrt(2)i+6,z=-sqrt(2)i+6

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(3 z - 18)^{2} + 58 = 40

z = 2 i + 6, z = - 2 i + 6

Solution steps

(3 z - 18)^{2} + 58 = 40

Move

58

to the right side

(3 z - 18)^{2} = - 18

For

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

the solutions are

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

Solve

3 z - 18 = - 18 : z = 2 i + 6

Solve

3 z - 18 = - - 18 : z = - 2 i + 6

The solutions to the quadratic equation are:

z = 2 i + 6, z = - 2 i + 6

6^{2} + 4^{2}

x^{2} - 90 x + 1944 = 0

\frac{- 1}{2 11} + 11

- 2 + 3 m - 1 < 4

\frac{b ^{4}}{b ^{2}} c

What are the solutions to the equation (3z-18)^2+58=40 ?
The solutions to the equation (3z-18)^2+58=40 are z=sqrt(2)i+6,z=-sqrt(2)i+6
Find the zeros of (3z-18)^2+58=40
The zeros of (3z-18)^2+58=40 are z=sqrt(2)i+6,z=-sqrt(2)i+6