What are the solutions to the equation (3z-18)^2+42=-3 ?

The solutions to the equation (3z-18)^2+42=-3 are z=sqrt(5)i+6,z=-sqrt(5)i+6

Find the zeros of (3z-18)^2+42=-3

The zeros of (3z-18)^2+42=-3 are z=sqrt(5)i+6,z=-sqrt(5)i+6

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(3 z - 18)^{2} + 42 = - 3

z = 5 i + 6, z = - 5 i + 6

Solution steps

(3 z - 18)^{2} + 42 = - 3

Move

42

to the right side

(3 z - 18)^{2} = - 45

For

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

the solutions are

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

Solve

3 z - 18 = - 45 : z = 5 i + 6

Solve

3 z - 18 = - - 45 : z = - 5 i + 6

The solutions to the quadratic equation are:

z = 5 i + 6, z = - 5 i + 6

r^{2} = 8

4 x^{2} - 10 = 39

x^{2} - x + 6 = 8

x^{2} + 5 + 4 x + 7 = 9 x + 12

(x + 3) (x + 4) = 6 x

What are the solutions to the equation (3z-18)^2+42=-3 ?
The solutions to the equation (3z-18)^2+42=-3 are z=sqrt(5)i+6,z=-sqrt(5)i+6
Find the zeros of (3z-18)^2+42=-3
The zeros of (3z-18)^2+42=-3 are z=sqrt(5)i+6,z=-sqrt(5)i+6