What are the solutions to the equation (z+4)^2=3 ?

The solutions to the equation (z+4)^2=3 are z=sqrt(3)-4,z=-sqrt(3)-4

Find the zeros of (z+4)^2=3

The zeros of (z+4)^2=3 are z=sqrt(3)-4,z=-sqrt(3)-4

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(z + 4)^{2} = 3

z = 3 - 4, z = - 3 - 4

Solution steps

(z + 4)^{2} = 3

For

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

the solutions are

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

Solve

z + 4 = 3 : z = 3 - 4

Solve

z + 4 = - 3 : z = - 3 - 4

The solutions to the quadratic equation are:

z = 3 - 4, z = - 3 - 4

(64 v^{3})^{\frac{4}{3}}

a^{4} - 256

x^{9} - 1

- 7 \geq 13 - 5 x

\frac{3 !}{( 3 - 1 ) ! 1 !} = (0.1)^{1} \cdot (0.9)^{2}

What are the solutions to the equation (z+4)^2=3 ?
The solutions to the equation (z+4)^2=3 are z=sqrt(3)-4,z=-sqrt(3)-4
Find the zeros of (z+4)^2=3
The zeros of (z+4)^2=3 are z=sqrt(3)-4,z=-sqrt(3)-4