What are the solutions to the equation 15+(x-4)^2=7 ?

The solutions to the equation 15+(x-4)^2=7 are x=2sqrt(2)i+4,x=-2sqrt(2)i+4

Find the zeros of 15+(x-4)^2=7

The zeros of 15+(x-4)^2=7 are x=2sqrt(2)i+4,x=-2sqrt(2)i+4

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15 + (x - 4)^{2} = 7

x = 22 i + 4, x = - 22 i + 4

Solution steps

15 + (x - 4)^{2} = 7

Move

15

to the right side

(x - 4)^{2} = - 8

For

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

the solutions are

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

Solve

x - 4 = - 8 : x = 22 i + 4

Solve

x - 4 = - - 8 : x = - 22 i + 4

The solutions to the quadratic equation are:

x = 22 i + 4, x = - 22 i + 4

3 (x^{2} - 3) = 35 x + 19

6 v^{2} + 3 = - 2 v

20 x^{2} + 6 x - 8 = 0

2 a^{2} - 288 = 0

x (4 x - 19) = 30

What are the solutions to the equation 15+(x-4)^2=7 ?
The solutions to the equation 15+(x-4)^2=7 are x=2sqrt(2)i+4,x=-2sqrt(2)i+4
Find the zeros of 15+(x-4)^2=7
The zeros of 15+(x-4)^2=7 are x=2sqrt(2)i+4,x=-2sqrt(2)i+4