What are the solutions to the equation (2x+4)^2=-1 ?

The solutions to the equation (2x+4)^2=-1 are x=-2+i 1/2 ,x=-2-i 1/2

Find the zeros of (2x+4)^2=-1

The zeros of (2x+4)^2=-1 are x=-2+i 1/2 ,x=-2-i 1/2

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(2 x + 4)^{2} = - 1

x = - 2 + i \frac{1}{2}, x = - 2 - i \frac{1}{2}

Solution steps

(2 x + 4)^{2} = - 1

For

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

the solutions are

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

Solve

2 x + 4 = - 1 : x = - 2 + i \frac{1}{2}

Solve

2 x + 4 = - - 1 : x = - 2 - i \frac{1}{2}

The solutions to the quadratic equation are:

x = - 2 + i \frac{1}{2}, x = - 2 - i \frac{1}{2}

X = 0.5 (X - 20) + 30

5 x^{2} - 22 x - 15 = 0

- v^{4} + 2 v^{3} + 4 v^{2} - 8 v

\frac{11 !}{2 ! ( 11 - 2 ) !}

3 (x - 5) < 2 (2 x - 1)

What are the solutions to the equation (2x+4)^2=-1 ?
The solutions to the equation (2x+4)^2=-1 are x=-2+i 1/2 ,x=-2-i 1/2
Find the zeros of (2x+4)^2=-1
The zeros of (2x+4)^2=-1 are x=-2+i 1/2 ,x=-2-i 1/2