What are the solutions to the equation 4(x-6)^2=-32 ?

The solutions to the equation 4(x-6)^2=-32 are x=2sqrt(2)i+6,x=-2sqrt(2)i+6

Find the zeros of 4(x-6)^2=-32

The zeros of 4(x-6)^2=-32 are x=2sqrt(2)i+6,x=-2sqrt(2)i+6

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4 (x - 6)^{2} = - 32

x = 22 i + 6, x = - 22 i + 6

Solution steps

4 (x - 6)^{2} = - 32

Divide both sides by

4

(x - 6)^{2} = - 8

For

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

the solutions are

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

Solve

x - 6 = - 8 : x = 22 i + 6

Solve

x - 6 = - - 8 : x = - 22 i + 6

The solutions to the quadratic equation are:

x = 22 i + 6, x = - 22 i + 6

9 (x^{2} - 6) + 17 = 12

- 3 x^{2} - 8 = 0

16 x^{2} - 49 = 0

4 (10 - 3 x)^{2} - 4 = 5

(x + 8)^{2} = - 1

What are the solutions to the equation 4(x-6)^2=-32 ?
The solutions to the equation 4(x-6)^2=-32 are x=2sqrt(2)i+6,x=-2sqrt(2)i+6
Find the zeros of 4(x-6)^2=-32
The zeros of 4(x-6)^2=-32 are x=2sqrt(2)i+6,x=-2sqrt(2)i+6