What are the solutions to the equation (z+1)^2+49=0 ?

The solutions to the equation (z+1)^2+49=0 are z=7i-1,z=-7i-1

Find the zeros of (z+1)^2+49=0

The zeros of (z+1)^2+49=0 are z=7i-1,z=-7i-1

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(z + 1)^{2} + 49 = 0

z = 7 i - 1, z = - 7 i - 1

Solution steps

(z + 1)^{2} + 49 = 0

Move

49

to the right side

(z + 1)^{2} = - 49

For

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

the solutions are

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

Solve

z + 1 = - 49 : z = 7 i - 1

Solve

z + 1 = - - 49 : z = - 7 i - 1

The solutions to the quadratic equation are:

z = 7 i - 1, z = - 7 i - 1

x < 3 x + 2 < 2 x + 6

(5 - 2 i) (5 + 2 i)

8 x^{2} + 12 = 7 x^{2} + 7 x

2 x^{3} - 11 x^{2} + 5 x

t^{2} - 7 t = 0

What are the solutions to the equation (z+1)^2+49=0 ?
The solutions to the equation (z+1)^2+49=0 are z=7i-1,z=-7i-1
Find the zeros of (z+1)^2+49=0
The zeros of (z+1)^2+49=0 are z=7i-1,z=-7i-1