What are the solutions to the equation (7z+2)^2+16=0 ?

The solutions to the equation (7z+2)^2+16=0 are z=-2/7+i 4/7 ,z=-2/7-i 4/7

Find the zeros of (7z+2)^2+16=0

The zeros of (7z+2)^2+16=0 are z=-2/7+i 4/7 ,z=-2/7-i 4/7

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(7 z + 2)^{2} + 16 = 0

z = - \frac{2}{7} + i \frac{4}{7}, z = - \frac{2}{7} - i \frac{4}{7}

Solution steps

(7 z + 2)^{2} + 16 = 0

Move

16

to the right side

(7 z + 2)^{2} = - 16

For

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

the solutions are

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

Solve

7 z + 2 = - 16 : z = - \frac{2}{7} + i \frac{4}{7}

Solve

7 z + 2 = - - 16 : z = - \frac{2}{7} - i \frac{4}{7}

The solutions to the quadratic equation are:

z = - \frac{2}{7} + i \frac{4}{7}, z = - \frac{2}{7} - i \frac{4}{7}

2 x^{2} - 8 x = 10

16 = 36 + x^{2} - 12 x - \frac{1}{2}

2^{x} = 32

∣ 2 x - 3 ∣ = ∣ x + 4 ∣

3^{7}

What are the solutions to the equation (7z+2)^2+16=0 ?
The solutions to the equation (7z+2)^2+16=0 are z=-2/7+i 4/7 ,z=-2/7-i 4/7
Find the zeros of (7z+2)^2+16=0
The zeros of (7z+2)^2+16=0 are z=-2/7+i 4/7 ,z=-2/7-i 4/7