What are the solutions to the equation (3z-18)^2+60=42 ?

The solutions to the equation (3z-18)^2+60=42 are z=sqrt(2)i+6,z=-sqrt(2)i+6

Find the zeros of (3z-18)^2+60=42

The zeros of (3z-18)^2+60=42 are z=sqrt(2)i+6,z=-sqrt(2)i+6

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(3 z - 18)^{2} + 60 = 42

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

Solution steps

(3 z - 18)^{2} + 60 = 42

Move

60

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

3 - 3 x^{2} = 5 x^{2} + 15

3 x^{2} - 41 x = - 60

x, h x^{2} + 9 h^{2} x = 3 + x

3 x^{2} - 64 = 0

h^{2} + 3 h = 30

What are the solutions to the equation (3z-18)^2+60=42 ?
The solutions to the equation (3z-18)^2+60=42 are z=sqrt(2)i+6,z=-sqrt(2)i+6
Find the zeros of (3z-18)^2+60=42
The zeros of (3z-18)^2+60=42 are z=sqrt(2)i+6,z=-sqrt(2)i+6